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When talking about the famous sum 1+2+3+..., we can define the zeta function $\zeta(s)=\sum_{n=1}^{\infty}n^{-s}$ for $Re(s)>1$. Then by doing analytic continuation, we can find the result is $\zeta(-1)=-\frac{1}{12}$.

So how about this sum: $\sum_{n=1}^{\infty}n^n$.

For example, I can define a function like this: $f(s)=\sum_{n=1}^{\infty}n^{-sn}$, and do the analytic continuation, and get the result for $f(-1)$. Is this possible? And if the result is not a closed form solution, how to get its numeric value?

user32688
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2 Answers2

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Underwood gave in $1946$, for $p> 2$,the bounds $$p^p\left(1+\frac{1}{4 (p-1)}\right)<\sum_{n=1}^{p}n^n<p^p\left(1+\frac{2}{e (p-1)}\right) $$

Claude Leibovici
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Try this method (Infinite series as integral representation). I don't think we have such advanced tools in math to get closed form but it should work as numerical method

$\displaystyle \sum_{k=x}^{\infty} f (k)=-2\pi\int_{-\infty}^{\infty} \frac {F(x-\frac {1}{2}+it)}{(e^{\pi t}+e^{-\pi t})^2}dt $

Wreior
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