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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.

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Another way:

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

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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.

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  • $\begingroup$ Thanks a lot! Well what should we do in case of when series is less than 1.8 I don't think this summing or identity will help $\endgroup$ – Lokesh Sangewar Jan 10 '17 at 15:11
  • $\begingroup$ Can this help? n/n+1<n+1/n+2 $\endgroup$ – Lokesh Sangewar Jan 10 '17 at 15:12
  • $\begingroup$ Yeah instead of 2 $\endgroup$ – Lokesh Sangewar Jan 10 '17 at 15:13
  • $\begingroup$ @LokeshSangewar IF other question THEN new post, please. $\endgroup$ – Did Jan 10 '17 at 15:13
  • $\begingroup$ Well I just need a hint for that not complete answer.. That's why not print a new one.... Well I'll try first $\endgroup$ – Lokesh Sangewar Jan 10 '17 at 15:14
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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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  • $\begingroup$ Well since it involves integration... It's kinda complex for me to understand $\endgroup$ – Lokesh Sangewar Jan 10 '17 at 15:19
  • $\begingroup$ @LokeshSangewar Is it easier now. $\endgroup$ – hamam_Abdallah Jan 10 '17 at 15:23
  • $\begingroup$ Yes.... It is. Lol $\endgroup$ – Lokesh Sangewar Jan 10 '17 at 15:25

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