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The infamous (false) result $$\sum_{n=1}^\infty n=-\frac1{12}$$ is due to Ramanujan. It's obtained by performing illegal manipulations on a divergent sum, basically treating it as if it were a real number. Now, by definition,

$$\zeta(s)=\sum_{n=1}^\infty\frac1{n^s}$$

when $\Re(s)>1$, otherwise we consider the analytic continuation (as the above series would diverge). However, if we were to apply such definition to the case $s=-1$, we would obtain

$$\zeta(-1)=\sum_{n=1}^\infty n$$

and, interestingly enough,

$$\zeta(-1)=-\frac1{12}$$

if we use the correct definition when $s=-1$. This cannot be a coincidence, however I don't see a clear connection between the two results.

Elvis
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  • That is the connection. By symbolically plugging in $s=-1$ into a series when it is not allowed gives you the infamous result that $\sum_{n=1}^{\infty}n=\frac{-1}{12}$. – peek-a-boo Feb 16 '25 at 01:32
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    The “infamous false” result is important in string theory. In physics analytic continuation is often used on divergent series. – Ghoster Feb 16 '25 at 01:34
  • @peek-a-boo I don't see it. – Elvis Feb 16 '25 at 01:39
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    Don’t see what? You just symbolically plug things in. It’s not justified, but you do it nevertheless and call it a day – peek-a-boo Feb 16 '25 at 01:40
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    Likewise, you could come up with nonsense like $-\frac{1}{2}=\zeta(0) ``=”\sum_{n=0}^{\infty}\frac{1}{n^0}=\sum_{n=0}^{\infty}1$. – peek-a-boo Feb 16 '25 at 01:43
  • Wikipedia: “many summation methods are used in mathematics to assign numerical values even to a divergent series” – Ghoster Feb 16 '25 at 01:44
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    Surely, the answer lies in the details of the "illegal manipulations" Ramanujan performed, so we first have to know what those were. – Gerry Myerson Feb 16 '25 at 01:49
  • @peek-a-boo Physicists use that one too. – Ghoster Feb 16 '25 at 01:52

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