Limit as n tends to infinity of $\frac{\sum_{i=1}^{n} \frac{i^2}{(i+1)\ln(i+1)}}{n^2}$

Question

I've been trying to compute the following limit for a while, I think it's $0$ but I can't find a way of proving it...

$$\lim_{n \to \infty}\frac{\sum_{i=1}^{n} \frac{i^2}{(i+1)\ln(i+1)}}{n^2}$$

I've tried to compare it, I've tried to simplify the expression of the sum to infinity, but nothing convincing came out... Could you hint me? Thank you!!!

Answer 1

Set $m=\lfloor\sqrt n\rfloor$. Since $\frac i{i+1}<1$, the numerator is less than \begin{align*} \sum_{i=1}^m & \frac{i}{\ln(i+1)} + \sum_{i=m+1}^n \frac{i}{\ln(i+1)} \\ &\le \sum_{i=1}^m \frac{m}{\ln(1+1)} + \sum_{i=m+1}^n \frac{n}{\ln(m+1)} \\ &= \frac{m^2}{\ln2} + \frac{n^2}{\ln(m+1)} \\ &\le \frac n{\ln2} + \frac{n^2}{\ln\sqrt n} = \frac n{\ln2} + \frac{2n^2}{\ln n}. \end{align*} This shows that your limit equals $0$.

Answer 2

Let $x_n=\frac{1}{2\log2}+\frac{4}{3\log3}+\cdots+\frac{n^2}{n\log n}$

$x_{n-1}=\frac{1}{2\log2}+\frac{4}{3\log3}+\cdots+\frac{(n-1)^2}{(n-1)\log (n-1)}$

$y_n=n^2$

By Stolz-Cesaro

$\lim_{n\to \infty}\frac{x_n-x_{n-1}}{y_n-y_{n-1}}=\lim_{n\to \infty}\frac{n^2/n\log n}{n^2-(n-1)^2}=\lim_{n\to \infty}\frac{n}{(2n-1)\log n}\le \lim_{n\to \infty}\frac{1}{2n-1}=0$ since $\log n<n$ for all $n$

Hence by comparison the limit is $0$.