The beginning of a series in a limit using integrals Question:
evaluate:
$\lim \limits_{n \to \infty} n\left(\dfrac 1{(n+2)^2}+\dfrac 1{(n+3)^2}+1\ldots+\dfrac 1{(2n+1)^2}\right)$
What we did
We found that the limit (using the integral of $\displaystyle \int \limits_1^2\dfrac 1{x}\mathrm dx$) is $0.5$.
But that is true only if the sum also includes $\dfrac 1 {(n+1)^2} $ as well. The question is - can you ignore the fact that element is missing since it's "the beginning" of an endless series? 
 A: $$\lim_{n\to\infty}n\sum_{r=2}^{n+1}\frac1{(n+r)^2}$$
$$=\lim_{n\to\infty}n\sum_{r=1}^n\frac1{(n+r)^2}+n\left(\frac1{(2n+1)^2}-\frac1{(n+1)^2}\right)$$
$$\lim_{n\to\infty}n\sum_{r=1}^n\frac1{(n+r)^2}=\lim_{n\to\infty}\frac1n\sum_{r=1}^n\frac1{\left(1+\dfrac rn\right)^2} $$  
$$\text{As }\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$
$$\lim_{n\to\infty}\frac1n\sum_{r=1}^n\frac1{\left(1+\dfrac rn\right)^2} =\int_0^1\frac1{(1+x)^2}$$
and $$\lim_{n\to\infty}n\left(\frac1{(2n+1)^2}-\frac1{(n+1)^2}\right)=\cdots=0$$
A: If you plot the graph of the function : $x \mapsto \frac{1}{x^2}$ you can easily be convinced by $$\int_{k}^{k+1} \frac{\text{d}x}{x^2} \leqslant \frac{1}{k^2} \leqslant \int_{k-1}^{k} \frac{\text{d}x}{x^2} $$ for all $k \geqslant 2$. Therefore, if you sum this from $k=n+2$ to $k=2n+1$ we got
$$ \int_{n+2}^{2n+2} \frac{\text{d}x}{x^2} \leqslant \sum_{k=n+2}^{2n+1} \frac{1}{k^2} \leqslant \int_{n+1}^{2n+1} \frac{\text{d}x}{x^2}$$
which is also
$$ \frac{1}{n+2} - \frac{1}{2n+2}\leqslant \sum_{k=n+2}^{2n+1} \frac{1}{k^2} \leqslant \frac{1}{n+1} - \frac{1}{2n+1}$$
Multiplying by $n \geqslant 0$,
$$ \frac{n}{n+2} - \frac{n}{2n+2}\leqslant n\sum_{k=n+2}^{2n+1} \frac{1}{k^2} \leqslant \frac{n}{n+1} - \frac{n}{2n+1}$$
And since $$ \lim_{n\to+\infty}\frac{n}{n+2} - \frac{n}{2n+2} = \frac12$$
and  $$ \lim_{n\to+\infty}\frac{n}{n+1} - \frac{n}{2n+1} = \frac12$$
The squeeze theorem tells us that 
$$\lim_{n\to+\infty}  n\sum_{k=n+2}^{2n+1} \frac{1}{k^2} = \frac 12$$
