Show that $\sum\limits_{n=1}^{32}\frac1{n^2}=1 + \frac{1}{4} +\frac{1}{9} +\dots+ \frac{1}{1024}<2$

Question

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.

Answer 1

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, $$1+\sum_{n=1}^{31}\frac{1}{(n+1)^2}\leq 2-\frac{1}{32}<2$$

Answer 2

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.

Answer 3

Another way:

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