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.

Another way:
$$\begin{align} \sum_{k=1}^{2^5} \frac{1}{k^2}
&\leq 1 + \sum_{k=2}^{2^5}\int_{k-1}^{k}\frac{1}{t^2} dt\\
&=1+\int_1^{2^5}\frac{1}{t^2} dt \\
&=2-\frac{1}{2^5}\\
&<2
\end{align}$$
