Problem of the Week #120 - July 14th, 2014

[Chris L T521]

Here's this week's problem!

-----

Problem: Show that $2\sqrt{3}\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n} = \pi$.

-----

Hint: [sp]Use the power series for $\arctan x$.[/sp]

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by Euge, Kiwi and MarkFL. You can find Mark's solution below.

[sp]Let:

\(\displaystyle S=2\sqrt{3}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n}=6\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)}\left(\frac{1}{\sqrt{3}}\right)^{2n+1}\)

Using the Maclaurin series for the inverse tangent function, we may now state:

\(\displaystyle S=6\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)=6\cdot\frac{\pi}{6}=\pi\)

Hence, we may conclude:

\(\displaystyle 2\sqrt{3}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n}=\pi\)[/sp]