Proof that $\sum_{k=2}^{\infty} \frac{H_k}{k(k-1)} $ where $H_n$ is the sequence of harmonic numbers converges?

Question

How to prove that $$\displaystyle \sum_{k=2}^{\infty} \dfrac{H_k}{k(k-1)} $$ where $H_n$ is the sequence of harmonic numbers converges and that $\dfrac{H_n}{n(n-1)}\to 0 \ $

I have already proven by induction that this equals $\left(2-\dfrac{1}{(n+1)}-\dfrac{h_{n+1}}{n} \right)$ for every $n\ge1$ but am not sure how to use this in solving my problem. Could anyone give me some tips?

Answer 1

The closed form expression for the partial sum $S_n=\sum_{k=2}^n H_k/(k(k-1)$ is $$S_n=2-\frac{1}{n}-\frac{H_n}{n}.$$ This can be shown by induction, or more easily by writing out the sum and using that $1/(k(k-1)=1/(k-1)-1/k$ to collect the contributions of $1,1/2,1/3,\cdots,1/n$ to the sum. As others have pointed out, $H_n-\ln(n)\to C$ where $C$ is Euler's constant, and from that one gets $H_n/n \to 0$ in a few steps, maybe using L'Hopital on $\ln n / n.$

So the limit of the partial sum $S_n$ is $2$, which is also the sum from $2$ to infinity of the terms $H_k/(k(k-1))$. That the $k^{th}$ term goes to zero now follows since the series converges, or alternately it may be shown directly using the generous upper bound $H_n \le n$, so that $H_n/(n(n-1) \le 1/(n-1) \to 0.$

Answer 2

This just begs to be telescoped:

$$\sum_{k=2}^\infty\sum_{n=1}^k \frac{1}{k(k-1)n} = \sum_{k=2}^\infty \frac{1}{k(k-1)} + \sum_{n=2}^\infty \frac{1}{n} \sum_{k=n}^\infty\frac{1}{k(k-1)}$$

$$= \sum_{k=2}^\infty (\frac{1}{k-1} -\frac{1}{k}) + \sum_{n=2}^\infty \frac{1}{n} \sum_{k=n}^\infty(\frac{1}{k-1} -\frac{1}{k})$$

$$= 1 + \sum_{n=2}^\infty \frac{1}{n}\cdot\frac{1}{n-1} =1 + \sum_{n=2}^\infty(\frac{1}{n-1}-\frac{1}{n}) =2$$