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I was able to prove by induction and also by using calculus that $\sum_{k=1}^{\infty}\frac{k^2}{2^k}=6 $, also that $\sum_{k=1}^{\infty}\frac{k^3}{3^k}=\frac{33}{8}=4.125$ and $\sum_{k=1}^{\infty}\frac{k^4}{4^k}=\frac{380}{81}= 4.691..$

Investigating some other values numerically (using WA) seems to indicate that $I(a)=\sum_{k=1}^{\infty}\frac{k^a}{a^k} $ is minimal (and about 4.11292518301340..) when $a=\pi$

But I can not find a proof. I tried $I'(a)=0$ but without any usable result. Is the assumption correct? Is there a proof?

Paul vdVeen
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    The result of the sum is a polylogarithm; see Wolfram|Alpha: https://www.wolframalpha.com/input?i=sum+k%5Ea%2Fa%5Ek+from+k%3D1+to+infinity Using Wolfram|Alpha to find the minimum suggests it's around $a=3.12009$. – Chris Lewis Jul 10 '23 at 22:13

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For a shortcut, use your three numbers and fit the equation of a parabola.

This would give $$f(x)=\frac{791 }{648}x^2-\frac{2585}{324}x+\frac{461}{27}$$ which is minimum for $$x_*=\frac{2585}{791} = 3.268$$ $$f(x_*)=\frac{2069399}{512568}=4.037$$ which is not too bad.

Claude Leibovici
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