Limit of $\lim_{n\rightarrow∞} \sum_{k=1}^{n}\frac{n}{(n+k)^2}$ What is the limit of $\lim_{n\rightarrow∞} \sum_{k=1}^{n}\frac{n}{(n+k)^2}$? I got lower bound by plugging  $k=n$ which results in limit equal to $1/4$. Similarly, the upper bound is 1, for $k=0$. I do not really know how to proceed, so I would appreciate a hint, I feel like I am missing something obvious.
 A: How to do it without integrals?  Here is one way.  But of course the method shown by Nevzat is much simpler.
Upper bound
\begin{align}
\sum_{k=1}^n\frac{n}{(n+k)^2} &< \sum_{k=1}^n\frac{n}{(n+k)(n+k-1)}
\\ &=
\sum_{k=1}^n \left[\frac{k}{n+k} - \frac{k-1}{n+k-1}\right]
\\ &=
\frac{n}{n+n} - \frac{0}{n+0}\qquad\text{(telescoping sum)}
\\ &= \frac{1}{2} .
\end{align}
Lower bound
\begin{align}
\sum_{k=1}^n\frac{n}{(n+k)^2} &> \sum_{k=1}^n\frac{n}{(n+k+1)(n+k)}
\\ &=
\sum_{k=1}^n \left[\frac{k+1}{n+k+1} - \frac{k}{n+k}\right]
\\ &=
\frac{n+1}{n+n+1} - \frac{1}{n+1}\qquad\text{(telescoping sum)}
\\ &= 
\frac{n^2}{(2n+1)(n+1)} .
\end{align}
So
$$
\frac{n^2}{(2n+1)(n+1)} < \sum_{k=1}^n\frac{n}{(n+k)^2} < \frac{1}{2} ,
\\
\lim_{n\to\infty}\frac{n^2}{(2n+1)(n+1)} 
\le \lim_{n\to\infty}\sum_{k=1}^n\frac{n}{(n+k)^2} 
\le \lim_{n\to\infty}\frac{1}{2} ,
\\
\frac{1}{2}
\le \lim_{n\to\infty}\sum_{k=1}^n\frac{n}{(n+k)^2} 
\le \frac{1}{2} .
$$
A: $$\lim_{n\to\infty} \sum_{k=1}^{n}\frac{n}{(n+k)^2}= \lim_{n\to\infty} \frac1n\sum_{k=1}^{n} \frac{1}{(1+\frac{k}{n})^2}=\int_{0}^1\frac{dx}{(1+x)^2}=\frac12$$
