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

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

I tried to look at $$ f_n(x) = \sum_{n=0}^\infty \frac{(-1)^n}{3n+1} x^n $$

And maybe taking it's derivative but it didn't work out well.

Any ideas?

Solution 1:

Hint

$$\sum_{n=0}^\infty\frac{(-1)^n}{3n+1}=\lim_{N\to\infty}\sum_{n=0}^N(-1)^n\int_0^1t^{3n}dt=\lim_{N\to\infty}\int_0^1\sum_{n=0}^N(-t^3)^ndt=\lim_{N\to\infty}\int_0^1\frac{1-(-t^3)^{N+1}}{1+t^3}dt$$

Now you need to prove two things:

• $$\int_0^1\frac{dt}{1+t^3}=\frac{\pi}{3\sqrt{3}}+\frac{\log 2}{3}$$
• $$\lim_{N\to\infty}\int_0^1\frac{(-t^3)^{N+1}}{1+t^3}dt=0$$

Solution 2:

For a geometric series.

$$\sum_{n=0}^{\infty} (-1)^n(x^n) = \frac{1}{1+x}$$

Substitute $x \rightarrow x^3$

$$\sum_{n=0}^{\infty} (-1)^n (x^{3n}) = \frac{1}{1+x^3}$$

Integrate the sides.

$$\sum_{n=0}^{\infty} (-1)^n\frac{x^{3n + 1}}{3n + 1} = \int \frac{dx}{1+x^3}$$

The hard part is the integration. Then let $x=1$