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

Question

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 its derivative but it didn't work out well.

Any ideas?

Answer 1

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.

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

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

Answer 2

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:

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

Answer 3

Provided below is an alternate solution that came to my mind first that involves complex numbers as opposed to integration.

Consider the Taylor series of the principal value of $\ln(1+z)$, which is:

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

Let $\omega=\exp(2\pi i/3)$ be a third root of unity. With this, we have:

$$ \begin{align*} \ln(1+1)&=\sum_{n=0}^\infty(-1)^n\left(\frac{1}{3n+1}-\frac{1}{3n+2}+\frac{1}{3n+3}\right) \\ \omega^2\ln(1+\omega)&=\sum_{n=0}^\infty(-1)^n\left(\frac{1}{3n+1}-\frac{\omega}{3n+2}+\frac{\omega^2}{3n+3}\right) \\ \omega\ln(1+\omega^2)&=\sum_{n=0}^\infty(-1)^n\left(\frac{1}{3n+1}-\frac{\omega^2}{3n+2}+\frac{\omega}{3n+3}\right) \\ \end{align*} $$ Since $1+\omega+\omega^2=0$, we have $S=\frac13(\ln 2 + \omega^2\ln(1+\omega)+\omega\ln(1+\omega^2))$. Note that $\ln z = \ln|z|+i\arg z$ if we're considering the principal value. We can now simplify this like so: $$ \begin{align*} S &= \frac13\left(\ln 2 + \omega^2\ln(1+\omega)+\omega\ln(1+\omega^2)\right) \\ &= \frac13\left(\ln 2 + \omega^2\ln(-\omega^2)+\omega\ln(-\omega)\right) \\ &= \frac13\left(\ln 2 + \overline{\omega\ln(-\omega)}+\omega\ln(-\omega)\right) \\ &= \frac13\left(\ln 2 + 2\mathrm{Re}[\omega\ln(-\omega)]\right) \\ &= \frac13\left(\ln 2 + 2\left(-\sin\left(\frac{2\pi}{3}\right)\arg(-\omega)\right)\right) \\ &= \frac13\left(\ln 2 + 2\left(-\frac{\sqrt3}{2}\cdot-\frac\pi3\right)\right) \\ &= \frac13\left(\ln 2 + \frac{\pi}{\sqrt3}\right) \\ \end{align*} $$

As a bonus, you can extend this logic to find a general expression for all odd coefficients of $n$, though it certainly isn't pretty. I'll leave what I got here, to save other people's time:

$$ \sum_{n=0}^\infty \frac{(-1)^n}{pn+1} = \int_0^1 \frac{dx}{x^p+1} = \frac{1}{p}\left(\ln 2 + 2\sum_{k=1}^{(p-1)/2}\left(\cos\frac{2\pi k}{p}\ln\left(2\cos\frac{\pi k}{p}\right)+\pi\sin\frac{2\pi k}{p}\left(\frac{k}{p}-\left\lceil\frac{k}{p}-\frac{1}{2}\right\rceil\right)\right)\right) $$