# Limit to This Summation?

So I solved https://brilliant.org/problems/summationnested-radical/ by Phani Ramadevu and it got me thinking: is there a limit to $$\large{\sum_{n=1}^\infty {\frac{n^m}{m^n}}}$$ as $$m$$ approaches infinity?

First few terms (found using W|A):

$$\sum_{n=1}^\infty {\frac{n^0}{0^n}}$$ is undefined,

$$\sum_{n=1}^\infty {\frac{n^1}{1^n}}$$ diverges,

$$\sum_{n=1}^\infty {\frac{n^2}{2^n}}=6$$,

$$\sum_{n=1}^\infty {\frac{n^3}{3^n}}=\frac{33}{8}=4.125$$,

$$\sum_{n=1}^\infty {\frac{n^4}{4^n}}=\frac{380}{81}\approx4.691$$,

$$\sum_{n=1}^\infty {\frac{n^5}{5^n}}=\frac{3535}{512}\approx6.904$$, and so forth.

Note by Alex Delhumeau
3 years ago

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If you're talking about the sequence $$\displaystyle f(m) = \sum_{n=1}^\infty \frac{n^m}{m^n}$$, then it diverges because the number of initial terms that have dominating values increases. Try it out yourself. You can easily see why! A more challenging part (perhaps tedious) is to determine the exact values of each of these $$f(m)$$.

- 3 years ago

That seems reasonable. I still think its curious that the series starts undefined, then becomes infinitely big before shrinking to 4.125 and then going off to infinity again. Pretty cool stuff.

Speaking of: that would make a good problem, "find the minimum of $$f(m)=\sum_{n=1}^\infty {\frac{n^{m}}{m^{n}}}$$. "

How would that be done analytically (w/o guess & check)?

- 3 years ago