Prove bounds of a strictly increasing sequence using integrals to approximate For the strictly increasing sequence $ \ x_n = \frac1{1^2} + \frac1{2^2} + \frac1{3^2} +\cdots+\frac1{n^2},$ for $n\ge1$.
(a) Prove the sequence is bounded above by $2$; deduce that is has a limit $L\le2$.
(b) Prove that the "tail" sequence $\{x_n\}$, $n\ge N$, is bounded below by $3/2$ for some number $N$.
*Don't just calculate $N$ explicitly, use geometric reasoning and calculus.
What I've tried:
(a) I'm not sure how to set up the integral for this. I know I have to find an area greater then $x_n$, and show it converges to $2$, but none of the boundaries I've tried worked.
(b) Typing this out I realized it is completely wrong, but this is what I have. We can estimate a lower bound using $\int_2^n \frac1{x^2} \, dx$.
As $n\rightarrow \infty$, $\int_2^n \frac1{x^2}\,dx\rightarrow \frac12$. Since this estimation excludes n=1, where $x_n=1$, we can add $1$ to include that term and get a lower bound of $3/2$.
 A: This should give the idea to prove (a) and (b).
$$\sum_{n=1}^\infty\frac1{n^2}<1+\sum_{n=2}^\infty\frac1{n^2-1}=1+\frac12\sum_{n=2}^\infty\left(\frac1{n-1}-\frac1{n+1}\right)=1+\frac12\cdot\frac32$$
For the latter step, note the telescopic series.
A: For the upper bound, notice that
$$1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \frac{1}{49} + \ldots < 1 + \frac{1}{4} + \frac{1}{4} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \ldots  = \\ 1 + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \ldots = \frac{7}{4}$$
For the lower bound, notice that 
$$1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \frac{1}{49} + \ldots > 1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{16} + \frac{1}{64} + \frac{1}{64} + \frac{1}{64} + \ldots  = \\ 1 + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \ldots = \frac{3}{2}$$
It was my intention to not write it down formally, I'm leaving that step to you. 
