Sum of 1

Sum of 1 DEFAULT

Sum of arithmetic series

Before I show you how to find the sum of arithmetic series, you need to know what an arithmetic series is or how to recognize it.

A series is an expression for the sum of the terms of a sequence. 

For example, 6 + 9 + 12 + 15 + 18 is a series for it is the expression for the sum of the terms of the sequence 6, 9, 12, 15, 18. 

By the same token, 1 + 2 + 3 + .....100 is a series for it is an expression for the sum of the terms of the sequence 1, 2, 3, ......100.

To find the sum of arithmetic series, we can start with an activity.

The arithmetic series formula will make sense if you understand this activity. Focus then a lot on this activity!

Sum of arithmetic series: How to find the sum of the sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.


Using the sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

Add the first and last terms of the sequence and write down the answer.

Then, add the second and next-to-last terms.

Continue with the pattern until there is nothing to add.

We get:

1 + 10 = 11

2 + 9 = 11

3 + 8 = 11

4 + 7 = 11

5 + 6 = 11

What patterns do see? The sum is always 11.

11 + 11 + 11 + 11 + 11 = 5 × 11 = 55

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55

As you can see instead of adding all the terms in the sequence, you can just do 5 × 11 since you will get the same answer.

Notice also that 5 × 11 =

10/2

× 11

Sum =

10/2

× (1 + 10)

We can make a generalizationthat will help us find the sum of arithmetic series.

Notice that 1 is the first term of the sequence. Notice also that 10 is the last term of the sequence.

Sum =

10/2

× (first term + last term)


10 is the number of terms in the sequence since the sequence has 10 terms.

Sum of arithmetic series formula

Sum of arithmetic series formula

Sum =

number of terms/2

× (first term + last term)

The following notation is more commonly used to find the sum of arithmetic series.

The sum Snof a1+ a2+ a3+ a4+ ... + an  is  Sn=  

n/2

× (a1+ an)

n is the number of term, a1is the first term, and anis the nth or last term.

You will have no problem now to find the sum of 1 + 2 + 3 + 4 + ... + 100.

n = 100, a1= 1, an= 100

Sn=  

100/2

× (1 + 100 )

Sn= 50 × 101 = 5050

Find the sum of the arithmetic series 5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50

n = 10, a1= 5, an= 50

Sn=  

10/2

× ( 5 + 50 )

Sn= 5 × 55 = 275

Observation:

5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50 = 5 × (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)

We already found the sum of 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 above. It is 55.

5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50 = 5 × 55 = 275

How to find the number of terms or n when looking the sum of arithmetic series


When looking for the sum of arithmetic series, it is not always easy to know the number of terms or n.

Just use the formula below to find n.

last term - first term /common difference

+ 1

The common difference is the samenumber that is added to each term

How many term here? 2 + 6 + 10 + 14 + ... + 78

Common difference is 4

Number of terms =  

78 - 2 /4

+ 1 = 19 + 1 = 20 terms

Summation Notation:

See the summation notation for the series 8 + 14 + 20 + 26 + 32 + 38. If you are having hard time to derive the explicit formula, review arithmetic sequence. The technique is explained in arithmetic sequence.

Summation notation

As you can see when

n = 1,  6 ×1 + 2 = 6 + 2 = 8

n = 2,   6 ×2 + 2 = 12 + 2 = 14

n = 3,   6 ×3 + 2 = 18 + 2 = 20

n = 4,   6 ×4 + 2 = 24 + 2 = 26

n = 5,   6 ×5 + 2 = 30 + 2 = 32

n = 6,   6 ×6 + 2 = 36 + 2 = 38

The big Greek letter that looks like an E is the Greek capital letter sigma. It is the equivalent of the English letter S for summation.

Find the summation notation for 5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50

A good observation may help you see that 5n is the explicit formula for 5, 10, 15, 20, 25, 30, 35, 40, 45, 50

Why? when n = 1, 5 × 1 = 5, when n = 2, 5 × 2 = 10, and so forth...

The upper limit is 10 since we have 10 terms

The lower limit is 1
$$ S_n = \sum_{i=1}^{10} 5n $$
Find the sum of arithmetic series with the quiz below:



Sours: https://www.basic-mathematics.com/sum-of-arithmetic-series.html

Summation

Addition of a finite sequence of numbers

This article is about sums of several elements. For more elementary aspects, see Addition. For infinite sums, see Series (mathematics). For other uses, see Summation (disambiguation).

In mathematics, summation is the addition of a sequence of any kind of numbers, called addends or summands; the result is their sum or total. Beside numbers, other types of values can be summed as well: functions, vectors, matrices, polynomials and, in general, elements of any type of mathematical objects on which an operation denoted "+" is defined.

Summations of infinite sequences are called series. They involve the concept of limit, and are not considered in this article.

The summation of an explicit sequence is denoted as a succession of additions. For example, summation of [1, 2, 4, 2] is denoted 1 + 2 + 4 + 2, and results in 9, that is, 1 + 2 + 4 + 2 = 9. Because addition is associative and commutative, there is no need of parentheses, and the result is the same irrespective of the order of the summands. Summation of a sequence of only one element results in this element itself. Summation of an empty sequence (a sequence with no elements), by convention, results in 0.

Very often, the elements of a sequence are defined, through regular pattern, as a function of their place in the sequence. For simple patterns, summation of long sequences may be represented with most summands replaced by ellipses. For example, summation of the first 100 natural numbers may be written as 1 + 2 + 3 + 4 + ⋯ + 99 + 100. Otherwise, summation is denoted by using Σ notation, where {\textstyle \sum } is an enlarged capital Greek lettersigma. For example, the sum of the first n natural numbers can be denoted as {\textstyle \sum _{i=1}^{n}i.}

For long summations, and summations of variable length (defined with ellipses or Σ notation), it is a common problem to find closed-form expressions for the result. For example,[a]

{\displaystyle \sum _{i=1}^{n}i={\frac {n(n+1)}{2}}.}

Although such formulas do not always exist, many summation formulas have been discovered—with some of the most common and elementary ones being listed in the remainder of this article.

Notation[edit]

Capital-sigma notation[edit]

Mathematical notation uses a symbol that compactly represents summation of many similar terms: the summation symbol, {\textstyle \sum }, an enlarged form of the upright capital Greek letter sigma. This is defined as

{\displaystyle \sum _{i\mathop {=} m}^{n}a_{i}=a_{m}+a_{m+1}+a_{m+2}+\cdots +a_{n-1}+a_{n}}

where i is the index of summation; ai is an indexed variable representing each term of the sum; m is the lower bound of summation, and n is the upper bound of summation. The "i = m" under the summation symbol means that the index i starts out equal to m. The index, i, is incremented by one for each successive term, stopping when i = n.[b]

This is read as "sum of ai, from i = m to n".

Here is an example showing the summation of squares:

{\displaystyle \sum _{i=3}^{6}i^{2}=3^{2}+4^{2}+5^{2}+6^{2}=86.}

In general, while any variable can be used as the index of summation (provided that no ambiguity is incurred), some of the most common ones include letters such as i, j, k, and n; the latter is also often used for the upper bound of a summation.

Alternatively, index and bounds of summation are sometimes omitted from the definition of summation if the context is sufficiently clear. This applies particularly when the index runs from 1 to n.[1] For example, one might write that:

{\displaystyle \sum a_{i}^{2}=\sum _{i=1}^{n}a_{i}^{2}.}

One often sees generalizations of this notation in which an arbitrary logical condition is supplied, and the sum is intended to be taken over all values satisfying the condition. For example:

{\displaystyle \sum _{0\leq k<100}f(k)}

is the sum of f(k) over all (integers) k in the specified range,

\sum _{x{\mathop {\in }}S}f(x)

is the sum of f(x) over all elements x in the set S, and

{\displaystyle \sum _{d\,|\,n}\;\mu (d)}

is the sum of \mu (d) over all positive integers ddividingn.[c]

There are also ways to generalize the use of many sigma signs. For example,

{\displaystyle \sum _{i,j}}

is the same as

{\displaystyle \sum _{i}\sum _{j}.}

A similar notation is applied when it comes to denoting the product of a sequence, which is similar to its summation, but which uses the multiplication operation instead of addition (and gives 1 for an empty sequence instead of 0). The same basic structure is used, with {\textstyle \prod }, an enlarged form of the Greek capital letter pi, replacing the {\textstyle \sum }.

Special cases[edit]

It is possible to sum fewer than 2 numbers:

  • If the summation has one summand x, then the evaluated sum is x.
  • If the summation has no summands, then the evaluated sum is zero, because zero is the identity for addition. This is known as the empty sum.

These degenerate cases are usually only used when the summation notation gives a degenerate result in a special case. For example, if n=m in the definition above, then there is only one term in the sum; if n=m-1, then there is none.

Formal definition[edit]

Summation may be defined recursively as follows:

{\displaystyle \sum _{i=a}^{b}g(i)=0}, for b < a;
\sum _{i=a}^{b}g(i)=g(b)+\sum _{i=a}^{b-1}g(i), for ba.

Measure theory notation[edit]

In the notation of measure and integration theory, a sum can be expressed as a definite integral,

\sum _{k{\mathop {=}}a}^{b}f(k)=\int _{[a,b]}f\,d\mu

where [a,b] is the subset of the integers from a to b, and where \mu is the counting measure.

Calculus of finite differences[edit]

Given a function f that is defined over the integers in the interval[m, n], the following equation holds:

{\displaystyle f(n)-f(m)=\sum _{i=m}^{n-1}(f(i+1)-f(i)).}

This is the analogue of the fundamental theorem of calculus in calculus of finite differences, which states that:

{\displaystyle f(n)-f(m)=\int _{m}^{n}f'(x)\,dx,}

where

{\displaystyle f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}}

is the derivative of f.

An example of application of the above equation is the following:

{\displaystyle n^{k}=\sum _{i=0}^{n-1}\left((i+1)^{k}-i^{k}\right).}

Using binomial theorem, this may be rewritten as:

{\displaystyle n^{k}=\sum _{i=0}^{n-1}\left(\sum _{j=0}^{i-1}{\binom {k}{j}}i^{j}\right).}

The above formula is more commonly used for inverting of the difference operator\Delta , defined by:

{\displaystyle \Delta (f)(n)=f(n+1)-f(n),}

where f is a function defined on the nonnegative integers. Thus, given such a function f, the problem is to compute the antidifference of f, a function {\displaystyle F=\Delta ^{-1}f} such that {\displaystyle \Delta F=f}. That is, {\displaystyle F(n+1)-F(n)=f(n).} This function is defined up to the addition of a constant, and may be chosen as[2]

{\displaystyle F(n)=\sum _{i=0}^{n-1}f(i).}

There is not always a closed-form expression for such a summation, but Faulhaber's formula provides a closed form in the case where {\displaystyle f(n)=n^{k}} and, by linearity, for every polynomial function of n.

Approximation by definite integrals[edit]

Many such approximations can be obtained by the following connection between sums and integrals, which holds for any increasing function f:

\int _{s=a-1}^{b}f(s)\ ds\leq \sum _{i=a}^{b}f(i)\leq \int _{s=a}^{b+1}f(s)\ ds.

and for any decreasing function f:

\int _{s=a}^{b+1}f(s)\ ds\leq \sum _{i=a}^{b}f(i)\leq \int _{s=a-1}^{b}f(s)\ ds.

For more general approximations, see the Euler–Maclaurin formula.

For summations in which the summand is given (or can be interpolated) by an integrable function of the index, the summation can be interpreted as a Riemann sum occurring in the definition of the corresponding definite integral. One can therefore expect that for instance

{\frac {b-a}{n}}\sum _{i=0}^{n-1}f\left(a+i{\frac {b-a}{n}}\right)\approx \int _{a}^{b}f(x)\ dx,

since the right hand side is by definition the limit for n\to \infty of the left hand side. However, for a given summation n is fixed, and little can be said about the error in the above approximation without additional assumptions about f: it is clear that for wildly oscillating functions the Riemann sum can be arbitrarily far from the Riemann integral.

Identities[edit]

The formulae below involve finite sums; for infinite summations or finite summations of expressions involving trigonometric functions or other transcendental functions, see list of mathematical series.

General identities[edit]

{\displaystyle \sum _{n=s}^{t}C\cdot f(n)=C\cdot \sum _{n=s}^{t}f(n)\quad } (distributivity)[3]
{\displaystyle \sum _{n=s}^{t}f(n)\pm \sum _{n=s}^{t}g(n)=\sum _{n=s}^{t}\left(f(n)\pm g(n)\right)\quad } (commutativity and associativity)[3]
{\displaystyle \sum _{n=s}^{t}f(n)=\sum _{n=s+p}^{t+p}f(n-p)\quad } (index shift)
{\displaystyle \sum _{n\in B}f(n)=\sum _{m\in A}f(\sigma (m)),\quad } for a bijectionσ from a finite set A onto a set B (index change); this generalizes the preceding formula.
{\displaystyle \sum _{n=s}^{t}f(n)=\sum _{n=s}^{j}f(n)+\sum _{n=j+1}^{t}f(n)\quad } (splitting a sum, using associativity)
{\displaystyle \sum _{n=a}^{b}f(n)=\sum _{n=0}^{b}f(n)-\sum _{n=0}^{a-1}f(n)\quad } (a variant of the preceding formula)
{\displaystyle \sum _{n=s}^{t}f(n)=\sum _{n=0}^{t-s}f(t-n)\quad } (the sum from the first term up to the last is equal to the sum from the last down to the first)
{\displaystyle \sum _{n=0}^{t}f(n)=\sum _{n=0}^{t}f(t-n)\quad } (a particular case of the formula above)
{\displaystyle \sum _{i=k_{0}}^{k_{1}}\sum _{j=l_{0}}^{l_{1}}a_{i,j}=\sum _{j=l_{0}}^{l_{1}}\sum _{i=k_{0}}^{k_{1}}a_{i,j}\quad } (commutativity and associativity, again)
{\displaystyle \sum _{k\leq j\leq i\leq n}a_{i,j}=\sum _{i=k}^{n}\sum _{j=k}^{i}a_{i,j}=\sum _{j=k}^{n}\sum _{i=j}^{n}a_{i,j}=\sum _{j=0}^{n-k}\sum _{i=k}^{n-j}a_{i+j,i}\quad } (another application of commutativity and associativity)
{\displaystyle \sum _{n=2s}^{2t+1}f(n)=\sum _{n=s}^{t}f(2n)+\sum _{n=s}^{t}f(2n+1)\quad } (splitting a sum into its odd and even parts, for even indexes)
{\displaystyle \sum _{n=2s+1}^{2t}f(n)=\sum _{n=s+1}^{t}f(2n)+\sum _{n=s+1}^{t}f(2n-1)\quad } (splitting a sum into its odd and even parts, for odd indexes)
{\displaystyle \left(\sum _{i=0}^{n}a_{i}\right)\left(\sum _{j=0}^{n}b_{j}\right)=\sum _{i=0}^{n}\sum _{j=0}^{n}a_{i}b_{j}\quad } (distributivity)
{\displaystyle \sum _{i=s}^{m}\sum _{j=t}^{n}{a_{i}}{c_{j}}=\left(\sum _{i=s}^{m}a_{i}\right)\left(\sum _{j=t}^{n}c_{j}\right)\quad } (distributivity allows factorization)
{\displaystyle \sum _{n=s}^{t}\log _{b}f(n)=\log _{b}\prod _{n=s}^{t}f(n)\quad } (the logarithm of a product is the sum of the logarithms of the factors)
{\displaystyle C^{\sum \limits _{n=s}^{t}f(n)}=\prod _{n=s}^{t}C^{f(n)}\quad } (the exponential of a sum is the product of the exponential of the summands)

Powers and logarithm of arithmetic progressions[edit]

{\displaystyle \sum _{i=1}^{n}c=nc\quad } for every c that does not depend on i
{\displaystyle \sum _{i=0}^{n}i=\sum _{i=1}^{n}i={\frac {n(n+1)}{2}}\qquad } (Sum of the simplest arithmetic progression, consisting of the first n natural numbers.)[2]: 52 
{\displaystyle \sum _{i=1}^{n}(2i-1)=n^{2}\qquad } (Sum of first odd natural numbers)
{\displaystyle \sum _{i=0}^{n}2i=n(n+1)\qquad } (Sum of first even natural numbers)
{\displaystyle \sum _{i=1}^{n}\log i=\log n!\qquad } (A sum of logarithms is the logarithm of the product)
{\displaystyle \sum _{i=0}^{n}i^{2}=\sum _{i=1}^{n}i^{2}={\frac {n(n+1)(2n+1)}{6}}={\frac {n^{3}}{3}}+{\frac {n^{2}}{2}}+{\frac {n}{6}}\qquad } (Sum of the first squares, see square pyramidal number.) [2]: 52 
{\displaystyle \sum _{i=0}^{n}i^{3}=\left(\sum _{i=0}^{n}i\right)^{2}=\left({\frac {n(n+1)}{2}}\right)^{2}={\frac {n^{4}}{4}}+{\frac {n^{3}}{2}}+{\frac {n^{2}}{4}}\qquad } (Nicomachus's theorem) [2]: 52 

More generally, one has Faulhaber's formula for p>1

{\displaystyle \sum _{k=1}^{n}k^{p}={\frac {n^{p+1}}{p+1}}+{\frac {1}{2}}n^{p}+\sum _{k=2}^{p}{\binom {p}{k}}{\frac {B_{k}}{p-k+1}}\,n^{p-k+1},}

where B_{k} denotes a Bernoulli number, and {\displaystyle {\binom {p}{k}}} is a binomial coefficient.

Summation index in exponents[edit]

In the following summations, a is assumed to be different from 1.

\sum _{i=0}^{n-1}a^{i}={\frac {1-a^{n}}{1-a}} (sum of a geometric progression)
{\displaystyle \sum _{i=0}^{n-1}{\frac {1}{2^{i}}}=2-{\frac {1}{2^{n-1}}}} (special case for a = 1/2)
{\displaystyle \sum _{i=0}^{n-1}ia^{i}={\frac {a-na^{n}+(n-1)a^{n+1}}{(1-a)^{2}}}} (a times the derivative with respect to a of the geometric progression)
{\displaystyle {\begin{aligned}\sum _{i=0}^{n-1}\left(b+id\right)a^{i}&=b\sum _{i=0}^{n-1}a^{i}+d\sum _{i=0}^{n-1}ia^{i}\\&=b\left({\frac {1-a^{n}}{1-a}}\right)+d\left({\frac {a-na^{n}+(n-1)a^{n+1}}{(1-a)^{2}}}\right)\\&={\frac {b(1-a^{n})-(n-1)da^{n}}{1-a}}+{\frac {da(1-a^{n-1})}{(1-a)^{2}}}\end{aligned}}}
(sum of an arithmetico–geometric sequence)

Binomial coefficients and factorials[edit]

Main article: Binomial coefficient § Sums of the binomial coefficients

There exist very many summation identities involving binomial coefficients (a whole chapter of Concrete Mathematics is devoted to just the basic techniques). Some of the most basic ones are the following.

Involving the binomial theorem[edit]

{\displaystyle \sum _{i=0}^{n}{n \choose i}a^{n-i}b^{i}=(a+b)^{n},} the binomial theorem
{\displaystyle \sum _{i=0}^{n}{n \choose i}=2^{n},} the special case where a = b = 1
{\displaystyle \sum _{i=0}^{n}{n \choose i}p^{i}(1-p)^{n-i}=1}, the special case where p = a = 1 − b, which, for {\displaystyle 0\leq p\leq 1,} expresses the sum of the binomial distribution
{\displaystyle \sum _{i=0}^{n}i{n \choose i}=n(2^{n-1}),} the value at a = b = 1 of the derivative with respect to a of the binomial theorem
{\displaystyle \sum _{i=0}^{n}{\frac {n \choose i}{i+1}}={\frac {2^{n+1}-1}{n+1}},} the value at a = b = 1 of the antiderivative
Sours: https://en.wikipedia.org/wiki/Summation
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1 + 2 + 3 + 4 + ⋯

Divergent series

The infinite series whose terms are the natural numbers1 + 2 + 3 + 4 + ⋯ is a divergent series. The nth partial sum of the series is the triangular number

{\displaystyle \sum _{k=1}^{n}k={\frac {n(n+1)}{2}},}

which increases without bound as n goes to infinity. Because the sequence of partial sums fails to converge to a finite limit, the series does not have a sum.

Although the series seems at first sight not to have any meaningful value at all, it can be manipulated to yield a number of mathematically interesting results. For example, many summation methods are used in mathematics to assign numerical values even to a divergent series. In particular, the methods of zeta function regularization and Ramanujan summation assign the series a value of −1/12, which is expressed by a famous formula[2]

{\displaystyle 1+2+3+4+\cdots =-{\frac {1}{12}},}

where the left-hand side has to be interpreted as being the value obtained by using one of the aforementioned summation methods and not as the sum of an infinite series in its usual meaning. These methods have applications in other fields such as complex analysis, quantum field theory, and string theory.[3]

In a monograph on moonshine theory, Terry Gannon calls this equation "one of the most remarkable formulae in science".[4]

Partial sums[edit]

The first six triangular numbers

Main article: Triangular number

The partial sums of the series 1 + 2 + 3 + 4 + 5 + 6 + ⋯ are 1, 3, 6, 10, 15, etc. The nth partial sum is given by a simple formula:

{\displaystyle \sum _{k=1}^{n}k={\frac {n(n+1)}{2}}.}

This equation was known to the Pythagoreans as early as the sixth century BCE.[5] Numbers of this form are called triangular numbers, because they can be arranged as an equilateral triangle.

The infinite sequence of triangular numbers diverges to +∞, so by definition, the infinite series 1 + 2 + 3 + 4 + ⋯ also diverges to +∞. The divergence is a simple consequence of the form of the series: the terms do not approach zero, so the series diverges by the term test.

Summability[edit]

Among the classical divergent series, 1 + 2 + 3 + 4 + ⋯ is relatively difficult to manipulate into a finite value. Many summation methods are used to assign numerical values to divergent series, some more powerful than others. For example, Cesàro summation is a well-known method that sums Grandi's series, the mildly divergent series 1 − 1 + 1 − 1 + ⋯, to 1/2. Abel summation is a more powerful method that not only sums Grandi's series to 1/2, but also sums the trickier series 1 − 2 + 3 − 4 + ⋯ to 1/4.

Unlike the above series, 1 + 2 + 3 + 4 + ⋯ is not Cesàro summable nor Abel summable. Those methods work on oscillating divergent series, but they cannot produce a finite answer for a series that diverges to +∞. Most of the more elementary definitions of the sum of a divergent series are stable and linear, and any method that is both stable and linear cannot sum 1 + 2 + 3 + ⋯ to a finite value; see below. More advanced methods are required, such as zeta function regularization or Ramanujan summation. It is also possible to argue for the value of −+1/12 using some rough heuristics related to these methods.

Heuristics[edit]

Passage from Ramanujan's first notebook describing the "constant" of the series

Srinivasa Ramanujan presented two derivations of "1 + 2 + 3 + 4 + ⋯ = −+1/12" in chapter 8 of his first notebook.[7][8][9] The simpler, less rigorous derivation proceeds in two steps, as follows.

The first key insight is that the series of positive numbers 1 + 2 + 3 + 4 + ⋯ closely resembles the alternating series1 − 2 + 3 − 4 + ⋯. The latter series is also divergent, but it is much easier to work with; there are several classical methods that assign it a value, which have been explored since the 18th century.[10]

In order to transform the series 1 + 2 + 3 + 4 + ⋯ into 1 − 2 + 3 − 4 + ⋯, one can subtract 4 from the second term, 8 from the fourth term, 12 from the sixth term, and so on. The total amount to be subtracted is 4 + 8 + 12 + 16 + ⋯, which is 4 times the original series. These relationships can be expressed using algebra. Whatever the "sum" of the series might be, call it c = 1 + 2 + 3 + 4 + ⋯. Then multiply this equation by 4 and subtract the second equation from the first:

{\displaystyle {\begin{alignedat}{7}c={}&&1+2&&{}+3+4&&{}+5+6+\cdots \\4c={}&&4&&{}+8&&{}+12+\cdots \\c-4c={}&&1-2&&{}+3-4&&{}+5-6+\cdots \end{alignedat}}}

The second key insight is that the alternating series 1 − 2 + 3 − 4 + ⋯ is the formal power series expansion of the function 1/(1 + x)2 but with x defined as 1. Accordingly, Ramanujan writes

{\displaystyle -3c=1-2+3-4+\cdots ={\frac {1}{(1+1)^{2}}}={\frac {1}{4}}.}

Dividing both sides by −3, one gets c = −+1/12.

Generally speaking, it is incorrect to manipulate infinite series as if they were finite sums. For example, if zeroes are inserted into arbitrary positions of a divergent series, it is possible to arrive at results that are not self-consistent, let alone consistent with other methods. In particular, the step 4c = 0 + 4 + 0 + 8 + ⋯ is not justified by the additive identity law alone. For an extreme example, appending a single zero to the front of the series can lead to a different result. [1]

One way to remedy this situation, and to constrain the places where zeroes may be inserted, is to keep track of each term in the series by attaching a dependence on some function.[11] In the series 1 + 2 + 3 + 4 + ⋯, each term n is just a number. If the term n is promoted to a function n−s, where s is a complex variable, then one can ensure that only like terms are added. The resulting series may be manipulated in a more rigorous fashion, and the variable s can be set to −1 later. The implementation of this strategy is called zeta function regularization.

Zeta function regularization[edit]

Plot of ζ(s). For s > 1, the series converges and ζ(s) > 1. Analytic continuation around the pole at s = 1leads to a region of negative values, including ζ(−1) = −+1/12.

In zeta function regularization, the series {\textstyle \sum _{n=1}^{\infty }n} is replaced by the series {\textstyle \sum _{n=1}^{\infty }n^{-s}}. The latter series is an example of a Dirichlet series. When the real part of s is greater than 1, the Dirichlet series converges, and its sum is the Riemann zeta functionζ(s). On the other hand, the Dirichlet series diverges when the real part of s is less than or equal to 1, so, in particular, the series 1 + 2 + 3 + 4 + ⋯ that results from setting s = –1 does not converge. The benefit of introducing the Riemann zeta function is that it can be defined for other values of s by analytic continuation. One can then define the zeta-regularized sum of 1 + 2 + 3 + 4 + ⋯ to be ζ(−1).

From this point, there are a few ways to prove that ζ(−1) = −+1/12. One method, along the lines of Euler's reasoning,[12] uses the relationship between the Riemann zeta function and the Dirichlet eta functionη(s). The eta function is defined by an alternating Dirichlet series, so this method parallels the earlier heuristics. Where both Dirichlet series converge, one has the identities:

{\displaystyle {\begin{alignedat}{7}\zeta (s)&{}={}&1^{-s}+2^{-s}&&{}+3^{-s}+4^{-s}&&{}+5^{-s}+6^{-s}+\cdots &\\2\times 2^{-s}\zeta (s)&{}={}&2\times 2^{-s}&&{}+2\times 4^{-s}&&{}+2\times 6^{-s}+\cdots &\\\left(1-2^{1-s}\right)\zeta (s)&{}={}&1^{-s}-2^{-s}&&{}+3^{-s}-4^{-s}&&{}+5^{-s}-6^{-s}+\cdots &=\eta (s).\end{alignedat}}}

The identity {\displaystyle (1-2^{1-s})\zeta (s)=\eta (s)} continues to hold when both functions are extended by analytic continuation to include values of s for which the above series diverge. Substituting s = −1, one gets −3ζ(−1) = η(−1). Now, computing η(−1) is an easier task, as the eta function is equal to the Abel sum of its defining series,[13] which is a one-sided limit:

{\displaystyle -3\zeta (-1)=\eta (-1)=\lim _{x\to 1^{-}}\left(1-2x+3x^{2}-4x^{3}+\cdots \right)=\lim _{x\to 1^{-}}{\frac {1}{(1+x)^{2}}}={\frac {1}{4}}.}

Dividing both sides by −3, one gets ζ(−1) = −+1/12.

Cutoff regularization[edit]

A graph depicting the series with layered boxes

The series 1 + 2 + 3 + 4 + ⋯

A graph depicting the smoothed series with layered curving stripes

After smoothing

A graph showing a parabola that dips just below the y-axis
Asymptotic behavior of the smoothing. The y-intercept of the parabola is −+1/12.[1]

The method of regularization using a cutoff function can "smooth" the series to arrive at −+1/12. Smoothing is a conceptual bridge between zeta function regularization, with its reliance on complex analysis, and Ramanujan summation, with its shortcut to the Euler–Maclaurin formula. Instead, the method operates directly on conservative transformations of the series, using methods from real analysis.

The idea is to replace the ill-behaved discrete series {\displaystyle \textstyle \sum _{n=0}^{N}n} with a smoothed version

{\displaystyle \sum _{n=0}^{\infty }nf\left({\frac {n}{N}}\right),}

where f is a cutoff function with appropriate properties. The cutoff function must be normalized to f(0) = 1; this is a different normalization from the one used in differential equations. The cutoff function should have enough bounded derivatives to smooth out the wrinkles in the series, and it should decay to 0 faster than the series grows. For convenience, one may require that f is smooth, bounded, and compactly supported. One can then prove that this smoothed sum is asymptotic to −+1/12 + CN2, where C is a constant that depends on f. The constant term of the asymptotic expansion does not depend on f: it is necessarily the same value given by analytic continuation, −+1/12.[1]

Ramanujan summation[edit]

The Ramanujan sum of 1 + 2 + 3 + 4 + ⋯ is also −+1/12. Ramanujan wrote in his second letter to G. H. Hardy, dated 27 February 1913:

"Dear Sir, I am very much gratified on perusing your letter of the 8th February 1913. I was expecting a reply from you similar to the one which a Mathematics Professor at London wrote asking me to study carefully Bromwich's Infinite Series and not fall into the pitfalls of divergent series. … I told him that the sum of an infinite number of terms of the series: 1 + 2 + 3 + 4 + ⋯ = −+1/12 under my theory. If I tell you this you will at once point out to me the lunatic asylum as my goal. I dilate on this simply to convince you that you will not be able to follow my methods of proof if I indicate the lines on which I proceed in a single letter. …"[14]

Ramanujan summation is a method to isolate the constant term in the Euler–Maclaurin formula for the partial sums of a series. For a function f, the classical Ramanujan sum of the series {\displaystyle \textstyle \sum _{k=1}^{\infty }f(k)} is defined as

{\displaystyle c=-{\frac {1}{2}}f(0)-\sum _{k=1}^{\infty }{\frac {B_{2k}}{(2k)!}}f^{(2k-1)}(0),}

where f(2k−1) is the (2k − 1)-th derivative of f and B2k is the 2k-th Bernoulli number: B2 = 1/6, B4 = −+1/30, and so on. Setting f(x) = x, the first derivative of f is 1, and every other term vanishes, so[15]

{\displaystyle c=-{\frac {1}{6}}\times {\frac {1}{2!}}=-{\frac {1}{12}}.}

To avoid inconsistencies, the modern theory of Ramanujan summation requires that f is "regular" in the sense that the higher-order derivatives of f decay quickly enough for the remainder terms in the Euler–Maclaurin formula to tend to 0. Ramanujan tacitly assumed this property.[15] The regularity requirement prevents the use of Ramanujan summation upon spaced-out series like 0 + 2 + 0 + 4 + ⋯, because no regular function takes those values. Instead, such a series must be interpreted by zeta function regularization. For this reason, Hardy recommends "great caution" when applying the Ramanujan sums of known series to find the sums of related series.

Failure of stable linear summation methods[edit]

A summation method that is linear and stable cannot sum the series 1 + 2 + 3 + ⋯ to any finite value. (Stable means that adding a term to the beginning of the series increases the sum by the same amount.) This can be seen as follows. If

1 + 2 + 3 + ⋯ = x

then adding 0 to both sides gives

0 + 1 + 2 + ⋯ = 0 + x = x

by stability. By linearity, one may subtract the second equation from the first (subtracting each component of the second line from the first line in columns) to give

1 + 1 + 1 + ⋯ = xx = 0.

Adding 0 to both sides again gives

0 + 1 + 1 + 1 + ⋯ = 0,

and subtracting the last two series gives

1 + 0 + 0 + ⋯ = 0,

contradicting stability.

Therefore, every method that gives a finite value to the sum 1 + 2 + 3 + ⋯ is not stable or not linear.[17]

Physics[edit]

In bosonic string theory, the attempt is to compute the possible energy levels of a string, in particular, the lowest energy level. Speaking informally, each harmonic of the string can be viewed as a collection of D − 2 independent quantum harmonic oscillators, one for each transverse direction, where D is the dimension of spacetime. If the fundamental oscillation frequency is ω, then the energy in an oscillator contributing to the n-th harmonic is nħω/2. So using the divergent series, the sum over all harmonics is −ħω(D − 2)/24. Ultimately it is this fact, combined with the Goddard–Thorn theorem, which leads to bosonic string theory failing to be consistent in dimensions other than 26.[18]

The regularization of 1 + 2 + 3 + 4 + ⋯ is also involved in computing the Casimir force for a scalar field in one dimension.[19] An exponential cutoff function suffices to smooth the series, representing the fact that arbitrarily high-energy modes are not blocked by the conducting plates. The spatial symmetry of the problem is responsible for canceling the quadratic term of the expansion. All that is left is the constant term −1/12, and the negative sign of this result reflects the fact that the Casimir force is attractive.

A similar calculation is involved in three dimensions, using the Epstein zeta-function in place of the Riemann zeta function.[21]

History[edit]

It is unclear whether Leonhard Euler summed the series to −+1/12. According to Morris Kline, Euler's early work on divergent series relied on function expansions, from which he concluded 1 + 2 + 3 + 4 + ⋯ = ∞.[22] According to Raymond Ayoub, the fact that the divergent zeta series is not Abel-summable prevented Euler from using the zeta function as freely as the eta function, and he "could not have attached a meaning" to the series.[23] Other authors have credited Euler with the sum, suggesting that Euler would have extended the relationship between the zeta and eta functions to negative integers.[24][25][26] In the primary literature, the series 1 + 2 + 3 + 4 + ⋯ is mentioned in Euler's 1760 publication De seriebus divergentibus alongside the divergent geometric series 1 + 2 + 4 + 8 + ⋯. Euler hints that series of this type have finite, negative sums, and he explains what this means for geometric series, but he does not return to discuss 1 + 2 + 3 + 4 + ⋯. In the same publication, Euler writes that the sum of 1 + 1 + 1 + 1 + ⋯ is infinite.[27]

In popular media[edit]

David Leavitt's 2007 novel The Indian Clerk includes a scene where Hardy and Littlewood discuss the meaning of this series. They conclude that Ramanujan has rediscovered ζ(−1), and they take the "lunatic asylum" line in his second letter as a sign that Ramanujan is toying with them.[28]

Simon McBurney's 2007 play A Disappearing Number focuses on the series in the opening scene. The main character, Ruth, walks into a lecture hall and introduces the idea of a divergent series before proclaiming, "I'm going to show you something really thrilling", namely 1 + 2 + 3 + 4 + ⋯ = −+1/12. As Ruth launches into a derivation of the functional equation of the zeta function, another actor addresses the audience, admitting that they are actors: "But the mathematics is real. It's terrifying, but it's real."[29][30]

In January 2014, Numberphile produced a YouTube video on the series, which gathered over 1.5 million views in its first month.[31] The 8-minute video is narrated by Tony Padilla, a physicist at the University of Nottingham. Padilla begins with 1 − 1 + 1 − 1 + ⋯ and 1 − 2 + 3 − 4 + ⋯ and relates the latter to 1 + 2 + 3 + 4 + ⋯ using a term-by-term subtraction similar to Ramanujan's argument.[32] Numberphile also released a 21-minute version of the video featuring Nottingham physicist Ed Copeland, who describes in more detail how 1 − 2 + 3 − 4 + ⋯ = 1/4 as an Abel sum, and 1 + 2 + 3 + 4 + ⋯ = −+1/12 as ζ(−1).[33] After receiving complaints about the lack of rigour in the first video, Padilla also wrote an explanation on his webpage relating the manipulations in the video to identities between the analytic continuations of the relevant Dirichlet series.[34]

In The New York Times coverage of the Numberphile video, mathematician Edward Frenkel commented: "This calculation is one of the best-kept secrets in math. No one on the outside knows about it."[31]

Coverage of this topic in Smithsonian magazine describes the Numberphile video as misleading and notes that the interpretation of the sum as −+1/12 relies on a specialized meaning for the equals sign, from the techniques of analytic continuation, in which equals means is associated with.[35]

References[edit]

  1. ^ abcdTao, Terence (April 10, 2010), The Euler–Maclaurin formula, Bernoulli numbers, the zeta function, and real-variable analytic continuation, retrieved January 30, 2014.
  2. ^Lepowsky, J. (1999). "Vertex operator algebras and the zeta function". In Naihuan Jing and Kailash C. Misra (ed.). Recent Developments in Quantum Affine Algebras and Related Topics. Contemporary Mathematics. 248. pp. 327–340. arXiv:math/9909178. Bibcode:1999math......9178L..
  3. ^Tong, David (February 23, 2012). "String Theory". p. 28–48. arXiv:0908.0333 [hep-th].
  4. ^Gannon, Terry (April 2010), Moonshine Beyond the Monster: The Bridge Connecting Algebra, Modular Forms and Physics, Cambridge University Press, p. 140, ISBN .
  5. ^Pengelley, David J. (2002). "The bridge between the continuous and the discrete via original sources". In Otto Bekken; et al. (eds.). Study the Masters: The Abel-Fauvel Conference. National Center for Mathematics Education, University of Gothenburg, Sweden. p. 3. ISBN ..
  6. ^Ramanujan's Notebooks, retrieved January 26, 2014
  7. ^Abdi, Wazir Hasan (1992), Toils and triumphs of Srinivasa Ramanujan, the man and the mathematician, National, p. 41
  8. ^Berndt, Bruce C. (1985), Ramanujan's Notebooks: Part 1, Springer-Verlag, pp. 135–136
  9. ^Euler, Leonhard (2006). "Translation with notes of Euler's paper: Remarks on a beautiful relation between direct as well as reciprocal power series". Translated by Willis, Lucas; Osler, Thomas J. The Euler Archive. Retrieved 2007-03-22. Originally published as Euler, Leonhard (1768). "Remarques sur un beau rapport entre les séries des puissances tant directes que réciproques". Mémoires de l'Académie des Sciences de Berlin (in French). 17: 83–106.
  10. ^Promoting numbers to functions is identified as one of two broad classes of summation methods, including Abel and Borel summation, by Knopp, Konrad (1990) [1922]. Theory and Application of Infinite Series. Dover. pp. 475–476. ISBN .
  11. ^Stopple, Jeffrey (2003), A Primer of Analytic Number Theory: From Pythagoras to Riemann, p. 202, ISBN .
  12. ^Knopp, Konrad (1990) [1922]. Theory and Application of Infinite Series. Dover. pp. 490–492. ISBN .
  13. ^Berndt et al. p. 53.
  14. ^ abBerndt, Bruce C. (1985), Ramanujan's Notebooks: Part 1, Springer-Verlag, pp. 13, 134.
  15. ^Natiello, Mario A.; Solari, Hernan Gustavo (July 2015), "On the removal of infinities from divergent series", Philosophy of Mathematics Education Journal, 29: 1–11, hdl:11336/46148.
  16. ^Barbiellini, Bernardo (1987), "The Casimir effect in conformal field theories", Physics Letters B, 190 (1–2): 137–139, Bibcode:1987PhLB..190..137B, doi:10.1016/0370-2693(87)90854-9.
  17. ^See v:Quantum mechanics/Casimir effect in one dimension.[unreliable source?]
  18. ^Zeidler, Eberhard (2007), "Quantum Field Theory I: Basics in Mathematics and Physics: A Bridge between Mathematicians and Physicists", Quantum Field Theory I: Basics in Mathematics and Physics. A Bridge Between Mathematicians and Physicists, Springer: 305–306, Bibcode:2006qftb.book.....Z, ISBN .
  19. ^Kline, Morris (November 1983), "Euler and Infinite Series", Mathematics Magazine, 56 (5): 307–314, doi:10.2307/2690371, JSTOR 2690371.
  20. ^Ayoub, Raymond (December 1974), "Euler and the Zeta Function"(PDF), The American Mathematical Monthly, 81 (10): 1067–1086, doi:10.2307/2319041, JSTOR 2319041, retrieved February 14, 2014.
  21. ^Lefort, Jean, "Les séries divergentes chez Euler"(PDF), L'Ouvert (in French), IREM de Strasbourg (31): 15–25, archived from the original(PDF) on February 22, 2014, retrieved February 14, 2014.
  22. ^Kaneko, Masanobu; Kurokawa, Nobushige; Wakayama, Masato (2003), "A variation of Euler's approach to values of the Riemann zeta function"(PDF), Kyushu Journal of Mathematics, 57 (1): 175–192, arXiv:math/0206171, doi:10.2206/kyushujm.57.175, archived from the original(PDF) on 2014-02-02, retrieved January 31, 2014.
  23. ^Sondow, Jonathan (February 1994), "Analytic continuation of Riemann's zeta function and values at negative integers via Euler's transformation of series", Proceedings of the American Mathematical Society, 120 (4): 421–424, doi:10.1090/S0002-9939-1994-1172954-7, retrieved February 14, 2014.
  24. ^Barbeau, E. J.; Leah, P. J. (May 1976), "Euler's 1760 paper on divergent series", Historia Mathematica, 3 (2): 141–160, doi:10.1016/0315-0860(76)90030-6.
  25. ^Leavitt, David (2007), The Indian Clerk, Bloomsbury, pp. 61–62.
  26. ^Complicite (April 2012), A Disappearing Number, Oberon, ISBN .
  27. ^Thomas, Rachel (December 1, 2008), "A disappearing number", Plus, retrieved February 5, 2014.
  28. ^ abOverbye, Dennis (February 3, 2014), "In the End, It All Adds Up to –1/12", The New York Times, retrieved February 3, 2014.
  29. ^ASTOUNDING: 1 + 2 + 3 + 4 + 5 + ... = –1/12 on YouTube.
  30. ^Sum of Natural Numbers (second proof and extra footage) on YouTube.
  31. ^Padilla, Tony, What do we get if we sum all the natural numbers?, retrieved February 3, 2014.
  32. ^Schultz, Colin (2014-01-31). "The Great Debate Over Whether 1 + 2 + 3 + 4... + ∞ = −1/12". Smithsonian. Retrieved 2016-05-16.

Bibliography[edit]

Further reading[edit]

  • Zwiebach, Barton (2004). A First Course in String Theory. Cambridge UP. ISBN . See p. 293.
  • Elizalde, Emilio (2004). "Cosmology: Techniques and Applications". Proceedings of the II International Conference on Fundamental Interactions. arXiv:gr-qc/0409076. Bibcode:2004gr.qc.....9076E.
  • Watson, G. N. (April 1929), "Theorems stated by Ramanujan (VIII): Theorems on Divergent Series", Journal of the London Mathematical Society, 1, 4 (2): 82–86, doi:10.1112/jlms/s1-4.14.82

External links[edit]

  • Lamb E. (2014), "Does 1+2+3… Really Equal –1/12?", Scientific American Blogs.
  • This Week's Finds in Mathematical Physics (Week 124), (Week 126), (Week 147), (Week 213)
  • The Euler-Maclaurin formula, Bernoulli numbers, the zeta function, and real-variable analytic continuation by Terence Tao
  • A recursive evaluation of zeta of negative integers by Luboš Motl
  • Link to Numberphile video 1 + 2 + 3 + 4 + 5 + ... = –1/12
  • Divergent Series: why 1 + 2 + 3 + ⋯ = −1/12 by Brydon Cais from University of Arizona
Sours: https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF

Hi Jo,

The question you asked relates back to a famous mathematician, Gauss.  In elementary school in the late 1700’s, Gauss was asked to find the sum of the numbers from 1 to 100.  The question was assigned as “busy work” by the teacher, but Gauss found the answer rather quickly by discovering a pattern.  His observation was as follows:

1 + 2 + 3 + 4 + … + 98 + 99 + 100

Gauss noticed that if he was to split the numbers into two groups (1 to 50 and 51 to 100), he could add them together vertically to get a sum of 101.

1     + 2   + 3   + 4   + 5   + … + 48 + 49 + 50

100 + 99 + 98 + 97 + 96 + … + 53 + 52 + 51

1 + 100 = 101
2 + 99 = 101
3 + 98 = 101
.
.
.
48 + 53 = 101
49 + 52 = 101
50 + 51 = 101

Gauss realized then that his final total would be 50(101) = 5050.

The sequence of numbers (1, 2, 3, … , 100) is arithmetic and when we are looking for the sum of a sequence, we call it a series.  Thanks to Gauss, there is a special formula we can use to find the sum of a series:

       

S is the sum of the series and n is the number of terms in the series, in this case, 100.

Hope this helps!

There are other ways to solve this problem. You can, for example, memorize the formula

This is an arithmetic series, for which the formula is:
S = n[2a+(n-1)d]/2
where a is the first term, d is the difference between terms, and n is the number of terms.
For the sum of the first 100 whole numbers:
a = 1, d = 1, and n = 100
Therefore, sub into the formula:
S = 100[2(1)+(100-1)(1)]/2 = 100[101]/2 = 5050

You can also use special properties of the particular sequence you have.

An advantage of using Gauss' technique is that you don't have to memorize a formula, but what do you do if there are an odd number of terms to add so you can't split them into two groups, for example "what is the sum of the first 21 whole numbers?" Again we write the sequence "forwards and backwards" but using the entire sequence.

1   +  2  +  3 + ... + 19 + 20 + 21
21 + 20 + 19 + ... +   3 +   2 +  1

Now if you add vertically you get

22 + 22 + 22 + ... + 22 + 22 + 22 = 21(22) = 462

But this is twice the sum of the first 21 whole numbers so

1 + 2 + 3 + ... + 19 + 20 + 21 = 462/2 = 231

Natasha, Paul and Penny

Sours: http://mathcentral.uregina.ca/qq/database/qq.02.06/jo1.html

Of 1 sum

Sums

Examples for

Summation is the addition of a list, or sequence, of numbers. If the summation sequence contains an infinite number of terms, this is called a series. Sums and series are iterative operations that provide many useful and interesting results in the field of mathematics.

Compute a finite summation of a mathematical expression.

Compute an indexed sum:

Sum an incompletely specified sequence of terms:


Find the sum of an infinite number of terms.

Compute an infinite sum:

Sum a geometric series:

Compute a sum over all integers:

Compute an infinite sum (limits unspecified):

Sum an incompletely specified infinite series:

Sours: https://www.wolframalpha.com/examples/mathematics/calculus-and-analysis/sums/
Sum of first n natural numbers - Derivation of a formula

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