Home » Uncategorized » A telescoping arctan series

A telescoping arctan series

Evaluate the sum

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

Solution

First of all note that:

    \begin{align*} \sum_{n=1}^N \arctan \frac{2}{n^2} &=\sum_{n=1}^N \arctan (n+1)-\arctan(n-1) \\ &=-\frac{\pi}{4}+\arctan N+\arctan(N+1) \end{align*}

Hence letting N \rightarrow +\infty we have that

    \[\sum_{n=1}^{\infty} \arctan \frac{2}{n^2} = \frac{3 \pi}{4}\]

Read more

2 Comments

  1. Other interesting serieses of this form are:

    1. \displaystyle \sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n^2+n+4}}   = \arctan 2.

    Proof: We have successively:

        \begin{align*}  \sum_{n=1}^{\infty}\arctan{\frac{2}{n^2+n+4}} &= \sum_{n=1}^{\infty}\arctan \left ({\frac{2(n+1)-2n}{n(n+1)(1+\frac{4}{n(n+1)})}}  \right ) \\ &=\sum_{n=1}^{\infty}\arctan \left({\frac{\frac{2}{n}-\frac{2}{n+1}}{1+\frac{2} {n}\cdot\frac{2}{n+1}}}\right )\\ &=\sum_{n=1}^{\infty}\left (\arctan{\frac{2}{n}-\arctan{\frac{2}{n+1}}}  \right ) \\ &=\arctan 2 \end{align*}

    2. \displaystyle \sum_{n=1}^{\infty} \arctan \frac{2n}{n^4-n^2+1}  = \frac{\pi}{2}.

    Proof: We simply note that

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

    and

        \[\arctan \left( \frac{2n}{n^4 - n^2 + 1} \right) = \arctan \left( n^2+n \right) - \arctan \left( n^2-n \right)\]

    Thus, the series telescopes.

    3. \displaystyle \sum_{n=1}^\infty \arctan\left(\frac{1}{n^2+n+1}\right)  = \frac{\pi}{4}.

    Proof: The same as before note that the sum telescopes , since:

        \begin{align*} \arctan \left ( \frac{1}{n^2+n+1} \right ) &= \arctan \left ( \frac{\left ( n+1 \right )-n}{1+n\left ( n+1 \right )} \right ) \\ &= \arctan (n+1) - \arctan n \end{align*}

    The result follows.

Leave a comment

Who is Tolaso?

Find out more at his Encyclopedia Page.

Donate to Tolaso Network