# $\lim_{n\rightarrow \infty} ( \arctan\frac{1}{2} + \arctan \frac{1}{2.2^2} +…+ \arctan \frac {1}{2n^2})$

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

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

Therefore,

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

Can you perform the telescoping sum now?

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