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

Question

I'm having problem solving this. I had a look at Showing $ \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\frac{\pi}{3\sqrt{3}}$ , but it didn't really helped.

Answer 1

The finite sum $$\sum_{k=0}^n\frac{1}{(3k+1)(3k-1)}$$ Can be expressed in terms of the digamma function, $$=\frac{1}{6}\left(\psi^{(0)}(n+2/3)-\psi^{(0)}(n+4/3)+\psi^{(0)}(1/3)-\psi^{(0)}(-1/3)\right)$$ Where $$\psi^{(n)}(z):=(-1)^{n+1}n!\sum_{k=0}^\infty\frac{1}{(z+k)^{n+1}}$$

This can be obtained by splitting the denominator using partial fractions.

The digamma function has the known special values $$\psi^{(0)}(1/3)=-\gamma-\frac{\pi}{2\sqrt{3}}-\frac{3\log 3}{2} \\ \psi^{(0)}(-1/3)=3-\gamma+\frac{\pi}{2\sqrt 3}-\frac{3\log 3}{2}$$ These can be deduced from the reflection and addition properties of the polygamma.

As for the limit $$\lim_{n\to\infty} \psi^{(0)}(n+2/3)-\psi^{(0)}(n+4/3)$$ We can use the asymptotic expansion $$\psi^{(0)}(z)\asymp \ln z \\ \text{as}~z\to\infty$$ To conclude $$\lim_{n\to\infty} \psi^{(0)}(n+2/3)-\psi^{(0)}(n+4/3)=\lim_{n\to\infty}\ln(n+2/3)-\ln(n+4/3) \\ =\lim_{n\to\infty}\ln\left(\frac{n+2/3}{n+4/3}\right)=\ln(1)=0$$ Hence, $$\sum_{k=0}^\infty\frac{1}{(3k+1)(3k-1)}=\lim_{n\to\infty}\frac{1}{6}\left(\psi^{(0)}(n+2/3)-\psi^{(0)}(n+4/3)+\psi^{(0)}(1/3)-\psi^{(0)}(-1/3)\right) \\ =\frac{1}{6}\left(-\gamma-\frac{\pi}{2\sqrt{3}}-\frac{3\log 3}{2}-3+\gamma-\frac{\pi}{2\sqrt 3}+\frac{3\log 3}{2}\right) \\ =\frac{-1}{2}-\frac{\pi}{6\sqrt 3}\approx -0.802299894...$$

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If this answer seems too "artificial" to you, I rebut that these kind of problems are exactly what special functions such as $\psi^{(n)}$ are designed for.

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Edit: A reasonably painless way of getting the desired result.

We can show that $$\psi^{(0)}(-z)-\psi^{(0)}(z)=\frac{1}{z}+\pi\cot(\pi z)$$ Quite easily from the recurrence relation $$\psi^{(m)}(z+1)=\psi^{(m)}(z)+\frac{(-1)^mm!}{z^{m+1}}$$ And the reflection identity $$\psi^{(m)}(1-z)-\psi^{(m)}(z)=\pi\frac{\mathrm d^m}{\mathrm dz^m}\cot(\pi z)$$ I have posted a proof of this reflection identity somewhere on this site, but I can't find it.

You can use these formulae (exercise) to show $$\psi^{(0)}(1/3)-\psi^{(0)}(-1/3)=-3-\pi\cot(\pi/3)=-3-\frac{\pi}{\sqrt{3}}$$

Answer 2

We shall use a standard approach that applies contour integration. We will integrate the function $f(z)=\frac{\cot(\pi z)}{(3z+1)(3z-1)}$ over the closed contour $C_N$ on which $|z|=N+1/2$, $N\in \mathbb{N_{\ge0}}$.

We will make use of the facts that $|\cot(\pi z)|$ is bounded on $|z|=N+1/2$ and that $f$ has simple poles at $z=\pm1/3$ and $z=n$, for $n\in \mathbb{Z}$ with $|n|<N$. Proceeding we have

$$\begin{align} \oint_{C_N}f(z)\,dz&=\frac1{2\pi i}\oint_{|z|=N+1/2} \frac{\cot(\pi z)}{(3z+1)(3z-1)}\,dz\tag1\\\\ &= \sum_{n=-N}^N \text{Res}\left(\frac{\cot(\pi z)}{(3z+1)(3z-1)},z=n\right)\\\\ &+ \text{Res}\left(\frac{\cot(\pi z)}{(3z+1)(3z-1)},z=1/3\right)\\\\ &+ \text{Res}\left(\frac{\cot(\pi z)}{(3z+1)(3z-1)},z=-1/3\right)\\\\ &=\sum_{n=-N}^N \frac1{(3n+1)(3n-1)}+\frac{\pi}{3\sqrt3}\\\\ &=2\sum_{n=0}^N \frac1{(3n+1)(3n-1)}+1+\frac{\pi}{3\sqrt3} \end{align}$$

Letting $N\to\infty$, the contour integral in $(1)$ goes to $0$ and we find that

$$\sum_{n=0}^\infty \frac1{(3n+1)(3n-1)}=-\frac12-\frac{\pi\sqrt3}{18}$$

And we are done!