Show that $\sum\limits_{n=1}^{32}\frac1{n^2}=1 + \frac{1}{4} +\frac{1}{9} +\dots+ \frac{1}{1024}<2$ Show that 
$$1 + \frac{1}{4} +\frac{1}{9} +\dots+ \frac{1}{1024} <2$$
I know that the denominators are perfect squares starting from that of $1$ till $32$. Also I know about this identity 
$$\frac{1}{n(n+1)} > \frac{1}{(n+1)^2} > \frac{1}{(n+1)(n+2)}.$$
But I am not able to implement it
Please help me.

We know $$\frac{1}{n(n+1)} > \frac{1}{(n+1)^2} \Rightarrow \frac{1}{n} - \frac{1}{n+1} > \frac{1}{(n+1)^2}$$ Adding for $n =1,2,\cdots,31$, we get, $$\frac{1}{1} - \frac{1}{2} > \frac{1}{4}$$ $$\frac{1}{2}-\frac{1}{3} > \frac{1}{9}$$ $$\vdots$$ $$\frac{1}{31}-\frac{1}{32} > \frac{1}{1024}$$Adding gives us $$\frac{1}{4} +\frac{1}{9} +\cdots +\frac{1}{1024}< 1-\frac{1}{32}$$ $$\Rightarrow 1+\frac{1}{4}+ \cdots+\frac{1}{1024}< 2-\frac{1}{32} < 2$$ Hope it helps.
