This isn't a duplicate: I am looking for a sum or sums which are similar to $1+2+3+\cdots=-\frac{1}{12}$ but which have an answer (via analytic continuation) of $\pm \frac{1}{3}$ rather than $-\frac{1}{12}$. By similar I mean it is an infinite series of whole numbers.
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8multiply the present series by 4? – spaceisdarkgreen Apr 01 '17 at 16:52
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See this answer the result can be anything – reuns Apr 01 '17 at 16:56
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In a comment to my answer : Is $1+2+3+4+\cdots=-\frac{1}{12}$ the unique ''value'' of this series? ,
Sangchul Lee reported a result from physics SE that gives:
$$ \lim_{s\to 0^+}\left[\mbox{analitic continuation of}\sum_{n=1}^\infty\frac{n}{(n+\alpha)^s} \right]=\frac{\alpha^2}{2}-\frac{1}{12} $$
I've not verified thi result (but a proof is here: Can we use analytic continuation to obtain $\sum_{n=1}^\infty n = b, b\neq -\frac{1}{12}$) and you can find $\alpha$ such that the limits becomes $ \frac{1}{3}$, but the series is not a series of integer numbers. So, I suspect that we cannot find such a series of whole numbers.
Emilio Novati
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