Evaluate
$$ \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)3^n } $$
I can show very easily that this series converges using the alternating series test. By setting
$$b_n = \frac{1}{(2n+1)3^n} \ \ \ \ \ \ \ \Rightarrow b_n \leq b_{n+1}$$
and
$$ \lim_{n \to \infty} \frac{1}{(2n+1)3^n} = 0$$
However, what is the sum of the series? I can't find it. I tried to write it out term by term but I don't see any pattern.
Answer
Note that the power series of $\arctan(x) $ is
$$ \arctan(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1} $$
Also note that $$ \arctan(\frac{1}{\sqrt{3} }) = \frac{\pi}{6} $$
Therefore,
$$ \frac{\pi}{6} = \sum_{n=0}^{\infty} (-1)^n \frac{\big(\frac{1}{\sqrt{3}}\big)^{2n+1}}{2n+1} = \sum_{n=0}^{\infty} \frac{6}{\sqrt{3}} \frac{(-1)^n}{(2n+1)3^n} $$
Thus,
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)3^n} = \frac{\sqrt{3}\pi }{6} $$
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