TITLE: How to find following limit of sequence?
QUESTION [5 upvotes]: How to following limit
$$\lim _{n \rightarrow \infty}\left(\frac{1}{n^{2}}+\frac{2}{n^{2}}+\cdots+\frac{(n-1)}{n^{2}}\right)\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}\right). $$

My attempt:
\begin{align*}
L_n&=\left(\frac{1}{n^{2}}+\frac{2}{n^{2}}+\cdots+\frac{(n-1)}{n^{2}}\right)\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}\right).\\
&=\left(\frac{n-1}{2n}\right)\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}\right).\\
\end{align*}
By using camparison with
$$\frac{n}{2n}\le \left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}\right)\le \frac{n}{n+1}$$
I got $$\frac{1}{4}\le \lim_{n\to \infty}L_n\le \frac{1}{2}$$
But I need exact answer. But I could not get. Its answer given is $\ln \sqrt 2$.
Thank you very much for your valuable time.

REPLY [2 votes]: I'm not sure about my answer but please check out this below
for the first part of the limit
$$\lim_{n\to \infty}f(n)\times g(n) = \lim_{n\to \infty}f(n)\times\lim_{n\to \infty}g(n)$$
$$\lim_{n\to \infty}\left(\sum_{i =1}^{n-1}\frac i{n^2}\right)\left(\sum_{j =1}^n\frac {1}{n+j}\right) = \left(\lim_{n\to \infty}\sum_{i =1}^{n-1}\frac i{n^2}\right)\left(\lim_{n\to \infty}\sum_{j =1}^n\frac {1}{n+j}\right)$$
The first part of limit is as follows
$$\lim_{n\to \infty}\sum_{i=1}^{n-1}\frac i{n^2} = \lim_{n\to \infty}\frac 1{n^2}\sum_{i=1}^{n-1}i = \lim_{n\to \infty}\frac {n(n-1)}{2n^2} = \frac 1{2}$$

The second part of the limit is as:

$$\int_0^1\frac {dx}{x+1} = \left|\ln(x+1)\right|_0^1 = \ln2$$
The answer is $\frac 1{2}\ln2 = \ln\sqrt2$