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Calculate $$\lim_{n\rightarrow \infty} \left( \arctan\frac{1}{2} + \arctan \frac{1}{2.2^2} +....+ \arctan \frac {1}{2n^2}\right)$$

My answer: i know that $$ \sum_{n=1}^N \arctan \left( \frac{2}{n^2} \right) =\sum_{n=1}^N \arctan (n+1)-\arctan(n-1)$$

as Im not able To find the $\sum_{k=1}^{n} \arctan \frac {1}{2k^2}$

I need help,,,,,any hints /solution will be aprreciated

thanks in advance

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$$\frac{1}{2n^2}=\frac{2}{4n^2}=\frac{2}{1+4n^2-1}=\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}$$

Therefore,

$$\arctan \left( \frac{1}{2n^2}\right) = \arctan (2n+1) - \arctan(2n-1) $$

Can you perform the telescoping sum now?

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  • $\begingroup$ @Messififa Try to use the well known formula : $$\arctan \left( \frac{b-a}{1+ab} \right) = \arctan(b) -\arctan (a)$$ $\endgroup$ – Jaideep stands with Monica Jul 27 '18 at 3:28
  • $\begingroup$ ya i understand that but after that how can i used telescope sum? can u elaborate that telescope sum $\endgroup$ – Messi fifa Jul 27 '18 at 3:32
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    $\begingroup$ @Messififa I suggest you to write for maybe $N=4$ or $N=5$ and you will see. $\endgroup$ – Ovi Jul 27 '18 at 3:34
  • $\begingroup$ @JaideepKhare i got $\frac{ -\pi}{4}$ is its correct $\endgroup$ – Messi fifa Jul 27 '18 at 4:01
  • $\begingroup$ @Messififa Well it's almost correct. Correct answer is $\frac{\pi}{4}$. Just the sign is opposite. $\endgroup$ – Jaideep stands with Monica Jul 27 '18 at 15:43

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