<s>[INST] Proof that $\sum_{k=2}^{\infty} \frac{H_k}{k(k-1)} $ where $H_n$ is the sequence of harmonic numbers converges? 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?
 [/INST] This just begs to be telescoped:
$$\sum_{k=2}^\infty\sum_{n=1}^k \frac{1}{k(k-1)n} = \sum_{k=2}^\infty \frac{1}{k(k-1)} + \sum_{n=2}^\infty \frac{1}{n} \sum_{k=n}^\infty\frac{1}{k(k-1)}$$
$$=  \sum_{k=2}^\infty (\frac{1}{k-1} -\frac{1}{k}) + \sum_{n=2}^\infty \frac{1}{n} \sum_{k=n}^\infty(\frac{1}{k-1} -\frac{1}{k})$$
$$= 1 + \sum_{n=2}^\infty \frac{1}{n}\cdot\frac{1}{n-1} =1 + \sum_{n=2}^\infty(\frac{1}{n-1}-\frac{1}{n}) =2$$
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