<s>[INST] 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.
 [/INST] For $n>0$, we have
$$\frac{1}{(n+1)^2}\leq \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$
and by sum,
$$1+\sum_{n=1}^{31}\frac{1}{(n+1)^2}\leq 2-\frac{1}{32}<2$$
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