Evaluate the following inverse trigonometric summation:

Question

$$ \sum_{n=1}^{\infty} \arctan\left(\frac{1}{2n^{2}}\right) $$ I have tried to approach the problem by a telescopic series, $$ \tan ^{-1}\left(\frac{1}{1+2 n^{2}-1}\right) $$ but after factorizing the denominator like this, $$ \begin{array}{r} (n \sqrt{2}+1)(n \sqrt{2}-1) \\ =\left(2 n^{2}-1\right) \end{array} $$ and subtracting the factors, I am left with 2 as the difference which prohibits me from using $\arctan(a)-\arctan(b)$ formula, please help.

Answer 1

You should notice that $$\arctan{\left(\frac{1}{2n^2}\right)}=\arctan{\left(2n+1\right)}-\arctan{\left(2n-1\right)}$$ Therefore the series is telescopic and you're left with: $$-\frac{\pi}{4}+\frac{\pi}{2}=\boxed{\frac{\pi}{4}}$$

Answer 2

Let; $$S= \sum_{r=1}^{\infty}arctan \left(\frac{1}{2r^2}\right)$$

To telescope the series, we need to align the format to $arctan\left(\frac{a-b}{1+ab}\right)$,

$$S= \sum_{r=1}^{\infty}arctan \left(\frac{2}{1+4r^2-1}\right)$$

$$S= \sum_{r=1}^{\infty}arctan \left(\frac{2}{1+(2r-1)(2r+1)}\right)$$

$$S= \sum_{r=1}^{n} arctan (2r-1)-arctan(2r+1)$$

$$S=\frac{\pi}{4}-arctan(2n+1)$$

Now as n tends to $\infty$ you get; $$S=\frac{\pi}{4}$$