Finding limit via Sandwich Theorem: $\lim_{n\to\infty} n\sum_{n+1}^{2n} \frac{1}{i^2}$ Question:
Use the Sandwich Theorem to find $$\lim_{n\to ∞} n\sum_{n+1}^{2n} \frac{1}{i^2}$$
Appreciate any guidance. 
 A: Note that
$$\frac1{i} - \frac1{i+1} = \frac{1}{i(i+1)} \leqslant \frac{1}{i^2} \leqslant \frac{1}{i(i-1)} = \frac1{i-1} - \frac1{i}$$
Summing we get
$$\frac{n}{n+1}- \frac{n}{2n+1}  \leqslant n\sum_{i=n+1}^{2n}\frac{1}{i^2}\leqslant \frac{n}{n}- \frac{n}{2n},$$
and
$$\frac{n^2}{(n+1)(2n+1)}  \leqslant n\sum_{i=n+1}^{2n}\frac{1}{i^2}\leqslant \frac{1}{2}.$$
Now apply squeezing to find the limit $1/2$.
A: Observe that, $ x \mapsto \dfrac1{x^2}$ is decreasing, then
$$
\frac1{(i+1)^2} \leq\int_i^{i+1}\frac1{x^2}dx\leq \frac1{i^2},\qquad i=1,2,3,\cdots,
$$ summing from $i=n+1$ to $i=2n$,
$$
\sum_{k=n+1}^{2n}\frac1{i^2}+\frac1{(2n+1)^2}-\frac1{(n+1)^2}\leq\int_{n+1}^{2n+1}\frac1{x^2}dx\leq \sum_{k=n+1}^{2n}\frac1{i^2}.
$$ and, by the sandwich theorem,
$$
\lim_{n \to \infty} n\sum_{k=n+1}^{2n}\frac1{i^2}=\lim_{n \to \infty} n\int_{n+1}^{2n+1}\frac1{x^2}dx
$$
Since
$$
n\int_{n+1}^{2n+1}\frac1{x^2}dx=\frac{n}{n+1}-\frac12
$$ you obtain 

$$
\lim_{n \to \infty} n\sum_{k=n+1}^{2n}\frac1{i^2}=\frac12.
$$

