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This claim is false $\sum_{n=1}^{\infty}n=\sum_{n=1}^{\infty}n^{-(-1)}= \zeta(-1)=-1/12$.

The error is that we should

$\sum_{n=1}^{\infty}n=\sum_{n=1}^{\infty}(1/n ^1)^{-1}=(0)^{-1}$.

Am I correct? It's difficult to say that an infinite sum like that don't diverge and that sum of positive numbers can give negative number.

Grigory M
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user119231
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    $\zeta(s)=\sum_{n=1}^{\infty}n^{-s}$ is only true when $\text{Re}(s)>1$. To evaluate at $s=-1$ you would need to use the functional equation. Point being: the error you're making is with your second equality: $\sum_{n=1}^{\infty}n^{-(-1)}\neq \zeta(-1)$. – Ralph Mellish Jan 05 '14 at 14:55
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    Also, no: $\sum_{n=1}^{\infty}(1/n^1)^{-1}\neq (0)^{-1}$. It seems as though you're thinking that $\sum (\text{stuff})^{-1}=(\sum \text{stuff})^{-1}$, which is not true, and you seem to be be thinking that $\sum_{n=1}^{\infty}1/n=0$, which is not true (perhaps you're not thinking these things, but it is the only way I can interpret what you wrote above "i'm i correct?"). – Ralph Mellish Jan 05 '14 at 15:02
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    Related: proof of Riemann's functional equation – Grigory M Jan 05 '14 at 15:25
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    Also related: Values of the Riemann Zeta function and the Ramanujan Summation – Grigory M Jan 05 '14 at 15:32
  • It's difficult to say that an infinite sum like that don't diverge and that sum of positive numbers can give negative number. - Difficult ? Yes ! But impossible ? No ! ;-) – Lucian Jan 05 '14 at 16:10
  • Ralph Mellish's statement above is correct/is a good answer. But, for historical reasons: Ramanujan famously sent a letter to University College in London containing the expression: $1+2+3+\cdots+\infty = -\frac{1}{12}$. This is meaningless/mathematically wrong. But, it exhibits a shorthand resembling the zeta function, which is to be expected from someone who received no formal training in mathematics (and thus, wouldn't follow all conventions relating to definitions of functions, etc.). – apnorton Jan 05 '14 at 16:11
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    Dear anorton, «meaningless/mathematically wrong» is... an overstatement, perhaps. There are various notions of summability beside convergence of the sequence of partial sums. – Grigory M Jan 05 '14 at 17:36

2 Answers2

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Use the functional equation: $$\zeta(s)=2^s\pi^{-1+s}\Gamma(1-s)\sin\frac{\pi s}2\zeta(1-s)\;,\;\;s\neq1\implies$$

$$\zeta(-1)=\frac1{2\pi^2}\cdot1\cdot(-1)\zeta(2)=-\frac1{12}$$

DonAntonio
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Ralph Mellish said :

$\zeta(s)=\sum_{n=1}^{\infty}n^{-s}$ is only true when Re(s)>1. To evaluate at s=−1 you would need to use the functional equation. Point being: the error you're making is with your second equality: ∑∞n=1n−(−1)≠ζ(−1). Also, no: ∑∞n=1(1/n1)−1≠(0)−1. It seems as though you're thinking that ∑(stuff)−1=(∑stuff)−1, which is not true, and you seem to be be thinking that ∑∞n=11/n=0, which is not true (perhaps you're not thinking these things, but it is the only way I can interpret what you wrote above "i'm i correct?"). –

but he doesnt know that this trick is often used in string theory which is the final theory of everything

(this was posted as an answer in the place of a comment because have small inductive reputation)

user119235
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