Computing $\sum\limits_{n=1}^q {1\over (kn-1)(kn)(kn+1)}$ as a telescoping series, when $k\geqslant3$?

Question

One knows that $$S_1(q)=\sum_{n=1}^q {1\over (n-1)(n)(n+1)} = \sum_{n=1}^q\frac12\left({\frac{1}{(n-1)n}-\frac{1}{n(n+1)}}\right)$$ and the RHS can be easily telescoped. The same approach works for $$S_2(q)=\sum_{n=1}^q {1\over (2n-1)(2n)(2n+1)}$$ However, for $$S_3(q)=\sum_{n=1}^q {1\over (3n-1)(3n)(3n+1)}$$ it is impossible to telescope using the same method than in the two cases above. So:

How should $$S_k(q)=\sum_{n=1}^q {1\over (kn-1)(kn)(kn+1)}$$ where $k\geqslant3$ is an integer, be telescoped?

Answer 1

Partial fractions:

$$\frac1{(3n-1)3n(3n+1)}=\frac a{3n-1}+\frac b{3n}+\frac c{3n+1}\implies$$

$$1=3an(3n+1)+b(3n-1)(3n+1)+3cn(3n-1)$$

Now, attach values to $\;n\;$ to get the values $\;a,b,c\;$ :

$$\begin{align*}&n=0\implies&1=-b\\{}\\ &n=\frac13\implies&1=2a\implies a=\frac12\\{}\\ &n=-\frac13\implies&-1=-2c\implies c=\frac12\end{align*}$$

Thus:

$$\frac1{(3n-1)3n(3n+1)}=\frac12\left(\frac1{3n-1}-\frac2{3n}+\frac1{3n+1}\right)$$

Answer 2

As you have noted, in the case of $k=1$, we can use Partial Fractions: $$ \begin{align} \frac1{(n-1)n(n+1)} &=\frac{1/2}{n-1}-\frac1n+\frac{1/2}{n+1}\\ &=\frac12\left(\frac1{n-1}-\frac1n\right)-\frac12\left(\frac1n-\frac1{n+1}\right)\tag{1} \end{align} $$

---

The case for $k=2$ is not as easy $$ \frac1{(2n-1)2n(2n+1)} =\frac{1/2}{2n-1}-\frac1{2n}+\frac{1/2}{2n+1}\tag{2} $$ The partial sums are hard to evaluate, though the infinite sum can be computed as an alternating harmonic series: $$ \begin{align} \sum_{n=1}^\infty\frac1{(2n-1)2n(2n+1)} &=\sum_{n=1}^\infty\left(\frac{1/2}{2n-1}-\frac1{2n}+\frac{1/2}{2n+1}\right)\\ &=\sum_{n=1}^\infty\left(\frac{1/2}{2n-1}-\frac1{2n}\right)+\sum_{n=2}^\infty\left(\frac{1/2}{2n-1}\right)\\ &=\underbrace{\sum_{n=1}^\infty\left(\frac1{2n-1}-\frac1{2n}\right)}_{\text{alternating harmonic series}}-\frac12\\ &=\log(2)-\frac12\tag{3} \end{align} $$

---

For $k\gt2$, even the infinite sums are hard because the terms for different $k$ do not overlap. Something other than telescoping series needs to be used. $$ \begin{align} \sum_{n=1}^\infty\frac1{(kn-1)kn(kn+1)} &=\sum_{n=1}^\infty\left(\frac{1/2}{kn-1}-\frac1{kn}+\frac{1/2}{kn+1}\right)\\ &=-\frac1{2k}\sum_{n=1}^\infty\left(\frac1n-\frac1{n-1/k}\right)-\frac1{2k}\sum_{n=1}^\infty\left(\frac1n-\frac1{n+1/k}\right)\\ &=-\frac1{2k}\left(H_{-1/k}+H_{1/k}\right)\tag{4} \end{align} $$ where $H_x$ is the extended Harmonic number as described in this answer. That answer shows how to compute certain values, but more can be computed.

---

Extended Harmonic Numbers with Rational Arguments

In extension of the values derived in this answer, we can compute the value of $H_{p/q}$ for any rational $p/q$.

We will use that for $x\in[-2\pi,2\pi]$ $$ \log\left(1-e^{ix}\right) =\log\left(2\sin(x/2)\right)+ix/2-\pi i/2\operatorname{sgn}(\sin(x/2))\tag{5} $$ and $$ \frac1q\sum_{k=0}^{q-1}e^{2\pi ijk/q}=[j\equiv0\pmod{q}]\tag{6} $$ where $[\cdots]$ are Iverson brackets.

For $0\lt p\le q$, $$ \begin{align} H_{p/q} &=\sum_{n=1}^\infty\left(\frac1n-\frac1{n+p/q}\right)\\ &=q\sum_{n=1}^\infty\left(\frac1{qn}-\frac1{qn+p}\right)\\ &=\frac qp+q\sum_{j=1}^\infty\left(\vphantom{\frac1j}[j\equiv0\pmod{q}]-[j\equiv p\pmod{q}]\right)\frac1j\\ &=\frac qp+q\sum_{j=1}^\infty\frac1q\sum_{k=1}^{q-1}\left(e^{2\pi ijk/q}-e^{2\pi i(j-p)k/q}\right)\frac1j\\ &=\frac qp+q\sum_{j=1}^\infty\frac1q\sum_{k=1}^{q-1}e^{2\pi ijk/q}\left(1-e^{-2\pi ipk/q}\right)\frac1j\\ &=\frac qp-\sum_{k=1}^{q-1}\log\left(1-e^{2\pi ik/q}\right)\left(1-e^{-2\pi ikp/q}\right)\\ &=\frac qp-\sum_{k=1}^{q-1}\left[\scriptsize\log\left(2\sin\left(\frac{\pi k}q\right)\right)+i\left(\frac{\pi k}q-\frac\pi2\right)\right]\left[\scriptsize\left(1-\cos\left(\frac{2\pi kp}q\right)\right)+i\sin\left(\frac{2\pi kp}q\right)\right]\\ &=\bbox[5px,border:2px solid #C0A000]{\frac qp-\sum_{k=1}^{q-1}{\scriptsize\left[\log\left(2\sin\left(\frac{\pi k}q\right)\right)\left(1-\cos\left(\frac{2\pi kp}q\right)\right)+\left(\frac{\pi}2-\frac{\pi k}q\right)\sin\left(\frac{2\pi kp}q\right)\right]}}\tag{7} \end{align} $$ Because $(7)$ is valid for $0\lt p\le q$, we can use $$ H_{-1/k}=H_{(k-1)/k}-\frac{k}{k-1}\tag{8} $$ Thus, we get $$ \bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty\frac1{(kn-1)kn(kn+1)} =-\frac1{2k}\left(H_{(k-1)/k}+H_{1/k}-\frac{k}{k-1}\right)}\tag{9} $$ where each of the Harmonic numbers can be computed with $(7)$.

Here is a list of sums for the first $6$ values of $k$ $$ \begin{array}{c|c} k&\sum\limits_{n=1}^\infty\frac1{(kn-1)kn(kn+1)}\\\hline 1&\infty\text{, or $\frac14$ if we sum from $n=2$}\\ 2&-\frac12+\log(2)\\ 3&-\frac12+\frac12\log(3)\\ 4&-\frac12+\frac34\log(2)\\ 5&\scriptsize-\frac12-\frac12\log(2)+\frac{5+\sqrt5}{20}\log\left(5+\sqrt5\right)+\frac{5-\sqrt5}{20}\log\left(5-\sqrt5\right)\\ 6&-\frac12+\frac13\log(2)+\frac14\log(3) \end{array}\tag{10} $$ Mathematica implementation of $(7)$:

h[p_,q_] := q/p-Sum[(1-Cos[2Pi k p/q])Log[2Sin[Pi k/q]] + Sin[2Pi k p/q](1/2-k/q)Pi,{k,1,q-1}]