2
$\begingroup$

I am struggling to sum the following series: $$\sum_{n=1}^{\infty}\arctan\left(\frac{2}{n^2}\right).$$ I am not able to start the problem. I guess, any intial hint would be helpful for me. Thanks in advance.

$\endgroup$
2

1 Answer 1

4
$\begingroup$

Note that $$\arctan\left(\frac{2}{n^2}\right)=\arctan\left(\frac{(n+1)-(n-1)}{1+(n+1)(n-1)}\right)=\arctan(n+1)-\arctan(n-1).$$ Hence, for $N\geq 2$, $$\sum_{n=1}^N\arctan\left(\frac{2}{n^2}\right)=\arctan(N+1)+\arctan(N)-\arctan(1)-\arctan(0).$$ Can you take it from here?

$\endgroup$
2
  • $\begingroup$ Sir, I have a doubt. If we expand the above series, then it will look something like $tan^{-1}(2)-tan^{-1}(0)+tan^{-1}(3)-tan^{-1}(1)+tan^{-1}(4)-tan^{-1}(2)....+tan^{-1}(n+1)-tan^{-1}(n-1)$ which will give us $-\frac{\pi}{4}+\frac{π}{2}$ which is equal to $\frac{\pi}{4}$. But textbook is giving the answer $\frac{3\pi}{4}$. Am i doing something wrong while expanding the series. $\endgroup$
    – userNoOne
    Commented Feb 9, 2018 at 4:40
  • 1
    $\begingroup$ @RAHUlJHa Your textbook is correct, revise your expansion and you will find the correct result. $\endgroup$
    – Robert Z
    Commented Feb 9, 2018 at 7:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .