Limit of $(n+n^{1/2}+\cdots+n^{1/n})/n$

Asked Oct 14 '17 at 04:31

Active Oct 14 '17 at 04:31

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I'm pretty sure this has been asked here before. But perhaps due to high flexibility in formulating this problem I couldn't find any useful resources.

Obviously a lower bound is $2$. And, numerically testing it in WolframAlpha, I'm quite convinced the limit should just be $2$.

My first thought was to see it as a Riemann partial sum but it failed, at least in the equi-partition case.

Could anybody help? Thanks.

asked Oct 14 '17 at 04:31

Vim

13,640

Duplicate of https://math.stackexchange.com/a/2450093/148510 – RRL Oct 14 '17 at 04:35
@RRL: your duplicate is a duplicate! – Martin Argerami Oct 14 '17 at 04:37
@RRL thanks much! – Vim Oct 14 '17 at 04:37
@RRL technically finding a lower bound isn't that hard or meaningful, for example, $-1$ is also a lower bound. It's just this lower bound happens to be the lower bound. :) – Vim Oct 14 '17 at 04:48

Limit of $(n+n^{1/2}+\cdots+n^{1/n})/n$

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