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

[Chris L T521]

Here's this week's problem!

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

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

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[Chris L T521]

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

[sp]Let:

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

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

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

Hence, we may conclude:

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