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I have to solve the following limit:

$\lim_{n\rightarrow\infty} \left(\frac{\ln 2}{2}+\frac{\ln 3}{3}+\cdots + \frac{\ln n}{n} \right)^{\frac{1}{n}}$

I'm just curious if there is a simple way to solve it. I think I solved it by using some pretty unusual trick: I just considered the sum approximation under the radical by using an integral and got $\approx \frac{\ln^{2}n}{2}$. Then I simply applied Cauchy D'Alembert and got 1. Still thinking of a simple way.

2 Answers 2

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For each (sufficiently large) $n$, $1< \left(\frac{\ln 2}{2}+\frac{\ln 3}{3}+\cdots + \frac{\ln n}{n} \right)^{1/n}<(1+1+\cdots+1)^{1/n}=(n-1)^{1/n}

By using (or proving) $\lim\limits_{n\to\infty}n^{1/n}=1$ it is easy to finish. (By Cauchy--D'Alembert, I'm guessing you mean something like this.)

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    @Jo$n$as: Emm, ok2012-06-03
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We use $\log(t+1)\leq t$ for $t\geq 0$. We have $a_n:=\frac 1n\log \left(\sum_{k=1}^n\frac{\ln k}k\right)=\frac 1n\log \left(1+\sum_{k=1}^n\frac{\ln k}k-1\right)\leq \frac 1n\left(\sum_{k=1}^n\frac{\ln k}k-1\right).$ Since $\frac{\ln n}n\to 0$, the Cesaro means converge to $0$ (use $\varepsilon$) hence $\lim_{n\to +\infty}\frac 1n\log \left(\sum_{k=1}^n\frac{\ln k}k\right)=0$ and the limit of $e^{a_n}$ is $1$.