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.

#Calculus #Limits #Summation #Pleasehelp #InfiniteSeries

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Comments

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 - 6 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)?

Alex Delhumeau - 6 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$