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

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?

• The first line needs a sum on the right-hand-side. – Michael Burr Apr 15 '17 at 13:44
• Thank you for mentioning, any ideas to my questions? – MathEnthusiast Apr 15 '17 at 13:46
• Sure you would not be interested by the sum of the infinite series rather than its partial sums? – Did Apr 15 '17 at 14:15
• – lab bhattacharjee Apr 15 '17 at 14:24
• @labbhattacharjee ...None of which touches the case $k\geqslant3$. – Did Apr 15 '17 at 14:27

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:

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

• Yes, I did manage to complete until here, but how does it telescope? Sorry, may need more help. – MathEnthusiast Apr 15 '17 at 14:01
• Your proof for this part applies to every $k\geqslant3$, replacing each $3$ by $k$... – Did Apr 15 '17 at 14:02
• I know that, the question that I have right now is according to how telescoping series work, intermediate terms will cancel, right now I can't seem to find any intermediate terms to cancel out. I am just very bad. – MathEnthusiast Apr 15 '17 at 14:07
• @MathEnthusiast Sorry, I must have understood something different: in the above there is no cancellation at all, of course, as every denominator has a different residue modulo $\;3\;$ ... – DonAntonio Apr 15 '17 at 14:18
• Alright, it's ok, it might help, +1 – MathEnthusiast Apr 15 '17 at 14:20

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 $$0\lt p\lt q$$, \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 $$0\lt p\lt q$$, 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 $$k$$ $$\begin{array}{c|c} k&\sum\limits_{n=1}^\infty\frac1{(kn-1)kn(kn+1)}\\\hline 1&\infty\text{, or \frac14 if we sum from n=2}\\ 2&-\frac12+\log(2)\\ 3&-\frac12+\frac12\log(3)\\ 4&-\frac12+\frac34\log(2)\\ 5&\scriptsize-\frac12-\frac12\log(2)+\frac{5+\sqrt5}{20}\log\left(5+\sqrt5\right)+\frac{5-\sqrt5}{20}\log\left(5-\sqrt5\right)\\ 6&-\frac12+\frac13\log(2)+\frac14\log(3) \end{array}\tag{10}$$

• @Did: I was in the process of editing. – robjohn Apr 15 '17 at 14:15
• @Did Sometimes it is not "half baked answers" but rather that sometimes someone (say, I) thinks things are something more or less straightforward, and then after posting one discovers a little nuissance here o there that was overlooked...I think that also can be useful for people asking: the problems and efforts in attacking some problems. – DonAntonio Apr 15 '17 at 14:24
• @Did: my answer was not corrected from a "half-baked" answer, but amended to handle cases that needed special handling. I have amended the answer yet again. I hope that is okay. – robjohn Apr 16 '17 at 16:39
• it is not clear that the OP understands the case $k=2$. The series does not telescope, but the terms do cancel to a point to leave a common alternating series. For $k\gt2$, there is no cancellation among terms. – robjohn Apr 16 '17 at 17:45
• I was not using a "fastest gun" approach. I was trying to point out that only the $k=1$ case actually telescopes. The other cases require more. Partial fractions are part of the solution, but more is needed to actually get a closed form for the sum. The more that is needed was the point of my edits today. – robjohn Apr 16 '17 at 18:07