Methods for finding and guessing closed forms of infinite series

Question

I want to prove $\displaystyle\sum_{k \ge 0} \Big(\frac{1}{3k+1} - \frac{1}{3k+2}\Big) = \frac{\pi}{\sqrt{27}}$

The reason for this question is I was doing the integral $\displaystyle\int_0^{\infty} \frac{1}{1-x^3}$ and I proved that this was equal to $\displaystyle\sum_{k \ge 0} \Big(\frac{1}{3k+1} - \frac{1}{3k+2}\Big)$ by proving it was equal to $\displaystyle\int_0^1 \frac{1-x}{1-x^3}$ and then using the infinite series for $\dfrac{1}{1-x}$ and swapping the order of integration and summation to get that answer.

I tried for some while to write this in closed form but to no avail. So then I checked that the answer was mysteriously $\dfrac{\pi}{\sqrt{27}}$.

So my question mainly is that given the summation answer, how would you guess and prove it was equal to $\dfrac{\pi}{\sqrt{27}}$ given that you do not have access to calculators (I would appreciate if you could also point me towards some more general methods/resources to do this kind of stuff also, because i often times calculate integrals as an infinite sum but I can not find the closed form answer...)

Thank you for your help

Answer 1

These kind of sums can be evalauted with properties of the digamma function. For example, the difference between two digamma functions may be expressed as:

$$ \psi(x) - \psi(y) = (x-y) \sum_{j=0}^{\infty} \frac{1}{(j+x)(j+y)} \tag{1}$$

Hence

\begin{align*} S =& \sum_{k=0}^{\infty} \left(\frac{1}{3k+1} - \frac{1}{3k+2} \right)\\ =& \sum_{k=0}^{\infty} \frac{1}{(3k+1)(3k+2)}\\ =& \frac{1}{9}\sum_{k=0}^{\infty} \frac{1}{\left(k+\frac{1}{3}\right)\left(k+\frac{2}{3}\right)}\\ =& \frac{1}{3}\left[\psi\left(\frac{2}{3}\right) - \psi\left(\frac{1}{3}\right)\right] \end{align*}

There is a reflection formula for the digamma function

$$ \psi(1-x) = \psi(x) +\pi\cot(\pi x) $$

with $\displaystyle x= \frac{1}{3}$

$$ \psi\left(\frac{2}{3}\right) - \psi\left(\frac{1}{3}\right) = \pi\cot\left(\frac{\pi}{3}\right) = \frac{\pi}{\sqrt{3}} $$

Then

$$ S = \frac{1}{3}\left[\psi\left(\frac{2}{3}\right) - \psi\left(\frac{1}{3}\right)\right] = \frac{\pi}{3\sqrt{3}} $$

Edit 1: Note that the integral

$$\int_{0}^{\infty} \frac{1}{1-x^3} dx $$

diverges, only the integral

$$\int_{0}^{1} \frac{1-x}{1-x^3} dx $$

can be evaluated with this method.

Edit 2: There is also an easy way to evalue this integral without digamma properties

Since $\displaystyle 1+x+x^2= \frac{1-x^3}{1-x}$

$$\int_{0}^{1} \frac{1-x}{1-x^3} dx = \int_{0}^{1} \frac{1}{1+x+x^2} dx$$

Now note

$$ \frac{1}{1+x+x^2} = \Im \left[\frac{2}{\sqrt{3}\left(x+\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)}\right]$$

\begin{align*} \int_{0}^{1} \frac{1}{1+x+x^2} dx =& \frac{2}{\sqrt{3}} \Im \left[\int_{0}^{1} \frac{1}{\left(x+\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)} dx \right]\\ =& \frac{2}{\sqrt{3}} \Im\left[ \ln\left(\frac{3}{2}-i\frac{\sqrt{3}}{2}\right)- \ln\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)\right] \\ =& \frac{2}{\sqrt{3}}\Im\left[ \ln\left(\frac{3}{2}+i\frac{\sqrt{3}}{2}\right)\right] \\ =& \frac{2}{\sqrt{3}} \Im\left[\ln|\sqrt{3}|+i\arctan\left(\frac{\sqrt{3}}{3}\right) \right]\\ =& \frac{2}{\sqrt{3}} \arctan\left(\frac{\sqrt{3}}{3}\right)\\ =& \frac{\pi}{3\sqrt{3}} \end{align*}

Answer 2

Observe that

$$\sum_{k=0}^\infty \left(\frac1{3k+1} - \frac1{3k+2}\right) = \sum_{k=0}^\infty \left(\frac1{3k+1} + \frac{-1}{3k+2} + \frac0{3k+3}\right) = \frac2{\sqrt3} \sum_{n=1}^\infty \frac{\sin\left(\frac{2n\pi}3\right)}n$$

Consider a periodic function $f$ for which

$$\begin{cases}f(x) = \dfrac\pi2(1-x) & x\in[0,1] \\ f(x+1)=f(x) \end{cases}$$

whose Fourier series is

$$f(x) = \sum_{n=1}^\infty \frac{\sin(n\pi x)}n$$

and recovering your sum is trivial:

$$\sum_{k=0}^\infty \left(\frac1{3k+1} - \frac1{3k+2}\right) = \frac2{\sqrt3} f\left(\frac23\right) = \boxed{\frac\pi{3\sqrt3}}$$

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Alternatively, without using Fourier series,

$$\begin{align*} \frac2{\sqrt3}\sum_{n=1}^\infty \frac{\sin\left(\frac{2n\pi}3\right)}n &= \frac2{\sqrt3}\Im \sum_{n=1}^\infty \frac{\left(e^{i\tfrac{2\pi}3}\right)^n}n \\ &= \frac2{\sqrt3}\Im \left(-\log\left(1-e^{i\tfrac{2\pi}3}\right)\right) \\ &= \frac2{\sqrt3}\cot^{-1}\left(\sqrt3\right) = \boxed{\frac\pi{3\sqrt3}} \end{align*}$$