1

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
  • 17,146
  • 4
  • 82
  • 124
user119231
  • 11
  • 1
  • 4
    $\zeta(s)=\sum_{n=1}^{\infty}n^{-s}$ 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
  • 3
    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
  • 2
    Related: [proof of Riemann's functional equation](http://math.stackexchange.com/q/586870/) – Grigory M Jan 05 '14 at 15:25
  • 2
    Also related: [Values of the Riemann Zeta function and the Ramanujan Summation](http://math.stackexchange.com/q/38731/) – 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
  • 2
    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

1

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
  • 208,478
  • 17
  • 131
  • 283
0

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
  • 1