$\lim_{N\to\infty}\sum_{i=1}^N \frac{\log_e^i}{i} /(\log_e^N)^2 $ =

Question

The value of $$\lim_{N\to\infty}\frac{1}{{(\log N)}^2}\,\sum_{i=1}^N \,\frac{\log i}{i}$$ is

(A) 0

(B) 1

(C) 1/2

(D) None of this

Really stuck on this problem.Please help or give a hint.

Answer 1

For $f(x)=\frac{\log x}{x}$, we have $f'(x)=\frac{1-\log x}{x^2}<0$ for $x\geq 3$. Thus $f$ is decreasing on $[3,\infty)$. One can see that that $$\int_3^N\frac{\log x}{x}dx=\sum_{n=3}^{N-1}\int_n^{n+1}\frac{\log x}{x}dx\leq \sum_{n=3}^{N-1}\int_n^{n+1}\frac{\log n}{n}dx=\sum_{n=3}^{N-1}\frac{\log n}{n},$$ and $$\int_3^N\frac{\log x}{x}dx=\sum_{n=3}^{N-1}\int_n^{n+1}\frac{\log x}{x}dx\geq \sum_{n=3}^{N-1}\int_n^{n+1}\frac{\log (n+1)}{n+1}dx=\sum_{n=4}^{N}\frac{\log n}{n}.$$ On the other hand by direct calculation $$\int_3^N\frac{\log x}{x}dx=\frac{1}{2}(\log N)^2-\frac{1}{2}(\log 3)^2.$$ It follows that $$\lim_{N\to\infty}\frac{1}{(\log N)^2}\sum_{n=1}^N\frac{\log n}{n}=\frac{1}{2}.$$

Answer 2

By Cesàro-Stolz $$\lim_{N\to +\infty}\frac{\sum_{k=1}^{N}\frac{\log k}{k}}{\log^2 N}=\lim_{N\to +\infty}\frac{\frac{\log(N+1)}{N+1}}{(\log(N+1)+\log(N))\log\left(1+\frac{1}{N}\right)}=\frac{1}{2}\lim_{N\to +\infty}\frac{1}{(N+1)\log\left(1+\frac{1}{N}\right)} $$ equals $\frac{1}{2}$.