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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
2 years, 4 months 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 \(f(m) \).

Pi Han Goh - 2 years, 4 months ago

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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)?

Alex Delhumeau - 2 years, 4 months ago

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