# A series with infinitely many logarithms: $\lim_{n\to\infty} \left(\frac{\ln 2}2+\frac{\ln 3}3+\cdots + \frac{\ln n}n \right)^{\frac1n}$

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.

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

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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Yes. This is a really short/elementary proof. Nice! Thanks :-) – user 1618033 Jun 2 '12 at 9:17
@jonas: why community wiki – user9413 Jun 2 '12 at 9:18
@Jonas Meyer: it's clear. – user 1618033 Jun 2 '12 at 9:30
Perhaps I'm missing something, but I see you're using strict inequalities. This means that 1<1, which I don't think makes this proof valid. – Ray Jun 2 '12 at 15:36
@Ray: The inequalities all apply for each particular (big enough) number $n$. Strict inequality is true. For example, $n-1<n$ is true. But if $a_n<b_n<c_n$ for all (big enough) $n$, and if $\lim a_n = L=\lim c_n$, then $\lim b_n = L$. This is more generally true if $a_n\leq b_n\leq c_n$. Recall that strict inequalities of terms do not typically carry over into strict inequalities of limits. For another example, $1/n>0$ for all $n>0$, but $\lim\limits_{n\to\infty}1/n=0$. (This is related to the fact that closed intervals are closed, but open intervals are usually not.) – Jonas Meyer Jun 2 '12 at 18:41

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 $1$.

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