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## What is the sum of 1+2+3+4+5 . . . (forever)? 2

What is the sum of 1 + 2 + 3 + 4 . . . all the way to infinity?

-1/12

It is -1/12

Aakash

The solution seems a little vague - care to explain?

Mrinmay Dhar

There are a few ways to prove that ζ(−1) = −1/12. One method, along the lines of Euler's reasoning, uses the relationship between the Riemann zeta function and the Dirichlet eta function η(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:

\begin{alignat}{8} \zeta(s)&{}={}&1^{-s}+2^{-s}&&{}+3^{-s}+4^{-s}&&{}+5^{-s}+6^{-s}+\cdots& \ 2\cdot2^{-s}\zeta(s)&{}={}& 2\cdot2^{-s}&& {}+2\cdot4^{-s}&&{} +2\cdot6^{-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{alignat} The identity (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, which is a one-sided limit:

-3\zeta(-1)=\eta(-1)=\lim{x\nearrow 1}\left(1-2x+3x^2-4x^3+\cdots\right)=\lim{x\nearrow 1}\frac{1}{(1+x)^2}=\frac14 Dividing both sides by −3, one gets ζ(−1) = −1/12.

Dan McCullam

For an explanation in english, this video is great: https://www.youtube.com/watch?v=w-I6XTVZXww

nnygren

It is probably important to note that this is not really true.

The series does diverge to infinity as expected, but the Zeta function regularization is used to assign a "sum" to divergent functions and in this case the value is -1/12.

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