# Finding the sum for the following series

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.

## 2 Answers

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,

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

• How do you know to start with the power series of arctan(x)? Thank you very much. Jun 17, 2019 at 15:34

Consider first $$f(x) = \sum^\infty_{n=0} \frac{(-1)^nx^{2n}}{2n+1}$$ for $$\lvert x \rvert < 1$$. We can rewrite $$\frac{x^{2n+1}}{2n+1} = \int^x_0 t^{2n} dt.$$ Then $$f(x) = \sum^\infty_{n=0 } \int^x_0 (-1)^nt^{2n} dt = \sum^\infty_{n=0} \int^x_0 (-t^2)^ndt = \int^x_0\left(\sum^\infty_{n=0} ( -t^2)^n \right)dt.$$ Now in this range of integration, we have $$\lvert t^2 \rvert < 1$$ and so this is a geometric series. Can you take it from here?

EDIT: You would still need to justify the swapping of the sum and the integral, but you will likely have covered a theorem that accomplishes this.