Home » Uncategorized » Arctan series

Arctan series

Evaluate the series

    \[\mathcal{S} = \sum_{n=1}^{\infty} \arctan \frac{1}{n^2}\]


One can argue that the same technique used to evaluate the sum here can be used here as well. Unfortunately, this is not the case as the sum does not telescope. However the technique used here is the way to go.

First of all we note that

    \[\sum_{n=1}^{\infty} \arctan \frac{1}{n^2} = -\arg \prod _{n=1}^{\infty} \left ( 1 - \frac {i} {n^2}\right )\]

However, it is known that

(1)   \begin{equation*}  \sin \pi z =\pi z \prod _{n=1}^{\infty} \left ( 1 - \frac {z^2} {n^2} \right )  \end{equation*}

Let z= \frac {\sqrt {2}}{2} +  \frac {i\sqrt {2}}{2}. We note that z^2=i. Hence,

    \begin{align*} \sum_{n=1}^{\infty} \arctan \frac{1}{n^2} &= -\arg \prod _{n=1}^{\infty} \left ( 1 - \frac {i} {n^2}\right ) \\ &=-\arg \left (\frac {\sin \left ( \frac { \pi \sqrt {2}}{2} + \frac {i \pi \sqrt {2}}{2}\right ) } { \frac {\pi \sqrt {2}}{2} + \frac {i \pi\sqrt {2}}{2}} \right ) \end{align*}

which is now a matter of calculations. The sum is equal to

    \[\sum_{n=1}^{\infty} \arctan \frac{1}{n^2} = \frac{\pi }{4} -\arctan \left ( \cot \frac {\pi \sqrt 2}{2} \tanh \frac {\pi \sqrt 2}{2}\right )\]

Read more

Leave a comment

Who is Tolaso?

Find out more at his Encyclopedia Page.

Donate to Tolaso Network