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?

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\begin{align}
\sum_{k = 2}^{\infty}{H_{k} \over k\pars{k - 1}} & =
\sum_{k = 2}^{\infty}H_{k}\pars{\int_{0}^{1}x^{k - 1}\,\dd x}
\pars{\int_{0}^{1}y^{k - 2}\,\dd y}
\\[5mm] & =
\int_{0}^{1}\int_{0}^{1}\sum_{k = 2}^{\infty}H_{k}\,\pars{xy}^{k}\,
{\dd x\,\dd y \over xy^{2}}
\\[5mm] & =
\int_{0}^{1}\int_{0}^{1}\bracks{-\,{\ln\pars{1 - xy} \over 1 - xy} - xy}
{\dd x\,\dd y \over xy^{2}}
\\[5mm] & =
-\int_{0}^{1}{1 \over y^{2}}
\int_{0}^{y}\bracks{{\ln\pars{1 - x} \over \pars{1 - x}x} + 1}\dd x\,\dd y
\\[5mm] & =
-\int_{0}^{1}\bracks{{\ln\pars{1 - x} \over \pars{1 - x}x} + 1}
\int_{x}^{1}{\dd y \over y^{2}}\,\dd x
\\[5mm] & =
-\int_{0}^{1}\bracks{{\ln\pars{1 - x} \over x^{2}} + {1 - x \over x}}
\,\dd x = \bbx{\large 2}
\end{align}
