Let $\displaystyle f(m) =\sum_{n=1}^\infty\frac{n^m}{m^n}$ . Find the real value of $m$ for which its corresponding $f(m)$ is minimized.

#Calculus

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Comments

I don't have a proof yet, but I suspect that the minimum is at $m=\pi$ .

Dylan Pentland - 6 years ago

I'm pretty sure it falls between $3.11$ and $3.13$ .

Pi Han Goh - 6 years ago

Yes, I checked 3.12 and it appears to lower... oh well, that would have been a really cool minimum :(

I don't know if this is really helpful but this is equivalent to minimizing ${Li}_{-n}(\frac{1}{n})$ . (The polylogarithm function)

Dylan Pentland - 6 years ago

Wow, this seems to be a pretty interesting problem.....did you make any progress on this @Pi Han Goh sir??

Aaghaz Mahajan - 1 year ago

No, I don't.

Pi Han Goh - 1 year 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}`	$\boxed{123}$

Let f(m)=∑n=1∞nmmn\displaystyle f(m) =\sum_{n=1}^\infty\frac{n^m}{m^n} f(m)=n=1∑∞​mnnm​. Find the real value of mmm for which its corresponding f(m)f(m)f(m) is minimized.

Comments

Let $\displaystyle f(m) =\sum_{n=1}^\infty\frac{n^m}{m^n}$ . Find the real value of $m$ for which its corresponding $f(m)$ is minimized.