Show that $\lim_{n\to\infty}n\sum_{i=1}^n\frac{1}{(n+i)^2}=\frac{1}{2}.$ Show that $$\lim_{n\to\infty}n\sum_{i=1}^n\frac{1}{(n+i)^2}=\frac{1}{2}.$$
I know that I'm supposed to interpret this as a Riemann sum, but the leading $n$ is throwing me off. 
 A: We have $$\lim_{n\to\infty} \sum_{k=1}^{n} \frac{n}{(n+k)^2} = \lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} \frac{1}{\left(1+ \frac{k}{n} \right)^2} = \int_{0}^{1} \frac{1}{(1+x)^2} \, dx = \frac{1}{2}.$$
A: Since Sameer Kailasa and Marty Cohen just provided the answer with Riemann sum, let me provide another solution which applies for finite or infinite values of $n$.
$$\sum_{i=1}^n \frac 1 {(n+i)^2}=\psi ^{(1)}(n+1)-\psi ^{(1)}(2 n+1)\tag 1$$ where appears the first derivative of the digamma function (that is to say the trigamma function).
If you look here, you will see that $$\psi ^{(1)}(x)=\frac 1 x+\frac 1 {2x^2}+\sum_{k=1}^\infty \frac {B_{2 k}} {x^{2 k+1}}\tag 2$$ Using $(2)$ in $(1)$, we then have $$\sum_{i=1}^n \frac 1 {(n+i)^2}=\frac{1}{2 n}-\frac{3}{8 n^2}+\frac{7}{48
   n^3}+O\left(\frac{1}{n^4}\right)$$ which makes $$n\sum_{i=1}^n \frac 1 {(n+i)^2}=\frac{1}{2 }-\frac{3}{8 n}+\frac{7}{48
   n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit and how it is approached.
For $n=10$, the exact value is $\frac{502856614213805}{1083847519827072}\approx 0.463955$ while the above approximation leads to $\frac{2227}{4800}\approx 0.463958$.
A: $\begin{array}\\
\lim_{n\to \infty}n\sum_{k=1}^{n} \frac1{(n+k)^2}
&=\lim_{n\to \infty}n\sum_{k=1}^{n} \frac1{n^2}\frac1{(1+k/n)^2}\\
&=\lim_{n\to \infty}\frac1{n}\sum_{k=1}^{n} \frac1{(1+k/n)^2}\\
&\to \int_0^1\frac{dx}{(1+x)^2}\\
&=\int_1^2\frac{dx}{x^2}\\
&=\int_1^2 x^{-2}dx\\
&=\frac{-1}{x}\big |_1^2\\
&= -\frac12+1\\
&= \frac12\\
\end{array}
$
A: For $n\gt0$,
$$
n\sum_{k=1}^n\frac1{(n+k)(n+k+1)}\lt n\sum_{k=1}^n\frac1{(n+k)^2}\lt n\sum_{k=1}^n\frac1{(n+k)(n+k-1)}\tag{1}
$$
Using partial fractions, we get
$$
\begin{align}
n\sum_{k=1}^n\frac1{(n+k)(n+k+1)}
&=n\sum_{k=1}^n\left(\frac1{n+k}-\frac1{n+k+1}\right)\\
&=n\left(\frac1{n+1}-\frac1{2n+1}\right)\\
&=\frac{n^2}{(n+1)(2n+1)}\\
&=\frac12-\frac{3n+1}{(2n+2)(2n+1)}\tag{2}
\end{align}
$$
and
$$
\begin{align}
n\sum_{k=1}^n\frac1{(n+k)(n+k-1)}
&=n\sum_{k=1}^n\left(\frac1{n+k-1}-\frac1{n+k}\right)\\
&=n\left(\frac1n-\frac1{2n}\right)\\
&=\frac12\tag{3}
\end{align}
$$
Considering $(1)$, $(2)$, and $(3)$, the Squeeze Theorem says that
$$
\lim_{n\to\infty}n\sum_{k=1}^n\frac1{(n+k)^2}=\frac12\tag{4}
$$
