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$$\displaystyle\sum_{n=1}^\infty\sum_{m=2}^\infty \frac{1}{m^{mn}}=\sum_{n=1}^\infty\left(\frac{1}{2^{2n}}+\frac{1}{3^{3n}}+\frac{1}{4^{4n}}+\cdots\right) \tag{$\star$}$$

I have a strong suspicion that the above summation converges, although I'm not sure how to prove it. But I'm more interested in the precise value of $(\star)$, especially a nice representation (if possible).

Here's what I have so far:

\begin{align} \sum_{n=1}^\infty\left(\frac{1}{2^{2n}}+\frac{1}{3^{3n}}+\frac{1}{4^{4n}}+\cdots\right) & = \sum_{n=1}^\infty\frac{1}{2^{2n}}+\sum_{n=1}^\infty\frac{1}{3^{3n}}+\sum_{n=1}^\infty\frac{1}{4^{4n}}+\cdots \\ & = \sum_{n=1}^\infty\frac{1}{\left(2^{2}\right)^n}+\sum_{n=1}^\infty\frac{1}{\left(3^{3}\right)^n}+\sum_{n=1}^\infty\frac{1}{\left(4^{4}\right)^n}+\cdots \\ & = \frac{1}{2^2-1}+\frac{1}{3^3-1}+\frac{1}{4^4-1}+\cdots\\ & =\sum_{n=2}^\infty \frac{1}{n^n-1} \end{align}

So $(\star)=\displaystyle\sum_{n=2}^\infty \frac{1}{n^n-1}$. From here, all I know how to do is get an approximate value of the solution. How would you go about finding its exact value?

Wolfram Alpha Summation Calculator

StubbornAtom
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C. Melton
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  • I think you have already proved that the series converges, as it is true without any doubt for $\sum\frac1{n^n-1}$. – user Dec 05 '19 at 12:29
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    With what accuracy do you need to find the sum ()? Easy to see that $() \approx \frac{1}{e}$. – Witold Dec 05 '19 at 12:40
  • Proving convergence is easy: $n^n-1>2^n$ for $n>1$. –  Dec 05 '19 at 13:45
  • Is there some way of proving whether an exact representational form exists or does not exist without actually finding such a form? – C. Melton Dec 05 '19 at 14:35
  • What do you exactly mean by "an exact representational form"? You may introduce a function: $\Phi(x)=\sum_{n=2}^\infty\frac{x^n}{n^n-1}$. Then the value of your series will be $\phi\equiv\Phi(1)$. Is it satisfactory? – user Dec 05 '19 at 15:24

1 Answers1

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If you want more figures $$\displaystyle \sum_{n=2}^\infty \frac{1}{n^n-1}=0.37605925334160927467565605197313357724916639675$$ All the digits are obtained summing up to $31$.

An amazing approximation of it is $$\frac{76 \pi ^2+651 \pi-563 }{308 \pi ^2+481 \pi+1385}$$ which is in a relative error of $2.56\times 10^{-18}\text{ %}$.

Edit

Using exact arithmetic, I computed the value for $1000$ exact decimal places (if you want the number, tell me). The result is obtained summing up to $n=386$.

Whet looked interesting (at least to me) is that, negelcting the $-1$ in denominator $$a_n=\frac{1}{n^n}\implies \frac{a_{n+1}}{a_n}=\frac{1}{e n}-\frac{1}{2 e n^2}+O\left(\frac{1}{n^3}\right)$$ So, writing $$\sum_{n=2}^\infty \frac{1}{n^n-1}=\sum_{n=2}^{k-1} \frac{1}{n^n-1}+\sum_{n=k}^\infty \frac{1}{n^n-1}=\sum_{n=2}^{k-1} \frac{1}{n^n-1}+e^{\frac{1}{2 e}}I_0\left(\frac{1}{2 e}\right)\frac{ k }{k^k-1}$$ So, if we want the remainder to be less than $\epsilon$ we need $$k\sim\frac{\log \left(\frac{b}{\epsilon }\right)}{W\left(\log \left(\frac{b}{\epsilon}\right)\right)}\qquad \text{where} \qquad b=e^{\frac{1}{2 e}}I_0\left(\frac{1}{2 e}\right)$$ Applied to the case where $\epsilon =10^{-1000}$, this gives $k=386.55$.

Claude Leibovici
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  • Hi Claude. I am not impressed by this approximation (just like by the well known rational approximations of $\pi$), sorry. Because it requires $18$ digits to achieve an approximation correct to $20$ digits. –  Dec 05 '19 at 13:40
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    @YvesDaoust. Me neither ! Just amazed. Cheers :-) – Claude Leibovici Dec 05 '19 at 13:43
  • @ClaudeLeibovici A very interesting approximation indeed! Unfortunately it’s not exact but I nevertheless appreciate your stimulating contribution to this discussion. Thank you :) – C. Melton Dec 05 '19 at 14:31
  • @C.Melton. You are very welcome ! Inverse symbolic calculators do not find anything for this number. Cheers. – Claude Leibovici Dec 05 '19 at 15:05