Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

up vote 11 down vote accepted

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.)

share|improve this answer
    
Yes. This is a really short/elementary proof. Nice! Thanks :-) –  Chris's sis Jun 2 '12 at 9:17
1  
@jonas: why community wiki –  user9413 Jun 2 '12 at 9:18
    
@Jonas Meyer: it's clear. –  Chris's sis 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
show 3 more comments

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 $1$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.