Prove $\frac1n\sum_{k=1}^n\frac1{ak-1}=a\sum_{k=1}^\infty\frac1{ak-1}\cdot\frac1{a(n+k)-1}$

Question

I came across this somewhat old question where the OP provides an unproved specific case of this identity for $n\in\Bbb N$,

$$S_n = \frac1n \sum_{k=1}^n \frac1{ak-1} = a \sum_{k=1}^\infty \frac1{ak-1} \cdot \frac1{a(n+k)-1}$$

Question 1: How do we prove this?

I fail to see how one can rearrange the finite sum into an infinite series; it almost resembles invoking a substitution while integrating.

$$\frac1n \left(\frac13 + \frac17 + \frac1{11} + \cdots + \frac1{4n-1}\right) = \frac13\cdot\frac1{4n+3} + \frac17\cdot\frac1{4n+7} + \frac1{11}\cdot\frac1{4n+11} + \cdots$$

Mathematica evaluates both sums in terms of the digamma function,

$$S_n = \frac{\psi\left(n+1-\frac1a\right) - \psi\left(1-\frac1a\right)}{an}$$

The easy part is expanding into partial fractions to recover this digamma form after finding the latter form of $S_n$, since

$$\psi(n) = -\gamma + \sum_{k=1}^\infty \left(\frac1k - \frac1{n+k-1}\right)$$

where $\gamma$ is the Euler-Mascheroni constant.

---

A very similar sum appeared while I was evaluating a definite integral; I managed to show

$$I = \int_0^1 \log\left(1+x^2\right) \log \left(\frac{1+x}{1-x}\right) \, \frac{dx}x = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2} \sum_{k=1}^n \frac1{2k-1}$$

but did not know how to continue. The identity for $S_n$ seems like a promising lead.

Question 2: Can we generalize the proof for $S_n$ and extend the identity to integer $m>1$ to rewrite

$$S_{n,m} = \frac1{n^m} \sum_{k=1}^n \frac1{ak-1} \stackrel{?}= \sum_{k=1}^\infty \frac1{ak-1} \cdot \boxed{f(a,m,n,k)}$$

It turns out that $I = \pi G - \frac74\zeta(3)$, so I think Q2's answer is affirmative in some cases.

Answer 1

For $a\ne0$ and $m\gt n$, $$ \begin{align} \sum_{k=1}^n\frac1{ak-1} &=\sum_{k=1}^m\frac1{ak-1}-\sum_{k=n+1}^m\frac1{ak-1}\tag{1a}\\ &=\sum_{k=1}^m\frac1{ak-1}-\sum_{k=n+1}^{m+n}\frac1{ak-1}+\sum_{k=m+1}^{m+n}\frac1{ak-1}\tag{1b}\\ &=\sum_{k=1}^m\frac1{ak-1}-\sum_{k=1}^m\frac1{a(k+n)-1}+\sum_{k=m+1}^{m+n}\frac1{ak-1}\tag{1c}\\ &=\sum_{k=1}^m\left(\frac1{ak-1}-\frac1{a(k+n)-1}\right) +\sum_{k=m+1}^{m+n}\frac1{ak-1}\tag{1d}\\ &=an\sum_{k=1}^\infty\frac1{(ak-1)(a(k+n)-1)}+\lim_{m\to\infty}\sum_{k=m+1}^{m+n}\frac1{ak-1}\tag{1e}\\ &=an\sum_{k=1}^\infty\frac1{(ak-1)(a(k+n)-1)}\tag{1f} \end{align} $$ Explanation:
$\text{(1a):}$ add and subtract $\sum\limits_{k=n+1}^m\frac1{ak-1}$
$\text{(1b):}$ add and subtract $\sum\limits_{k=m+1}^{m+n}\frac1{ak-1}$
$\text{(1c):}$ substitute $k\mapsto k+n$ in the middle sum
$\text{(1d):}$ combine the first two sums
$\text{(1e):}$ take the limit as $m\to\infty$
$\text{(1f):}$ evaluate the second limit

Answer 2

Consider the following expression of partial fractions:

$$\frac{1}{ak-1} - \frac{1}{a(n+k) - 1 } = \frac{ (a(n+k)-1 ) - (ak-1) } {(ak-1) \times (a(n+k)-1 ) } = an \times \frac{ 1}{ak-1} \times \frac{1}{a(n+k)-1}$$

Summing up both sides from $k=1$ to infinity (Check if/when the summation is valid! EG we need $ a \neq 0$), we get (Observe that the LHS telescopes, which is why we have a finite sum.)

$$ \sum_{k=1}^n \frac{1}{ak-1} = an \sum_{k=1}^\infty \frac{ 1}{ak-1} \times \frac{1}{a(n+k)-1}.$$