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?

$$\sum_{k=2}^{\infty} \frac{H_k}{k(k-1)}=\sum_{k=1}^{\infty} \frac{H_{k+1}}{k(k+1)}=-\int_0^1 \ln(1-x)\sum_{k=1}^\infty \frac{x^k}{k}dx$$
$$=\int_0^1\ln^2(1-x)dx=\int_0^1 \ln^2xdx=2$$
Note that we use $\int_0^1 x^k \ln(1-x)dx=-\frac{H_{k+1}}{k+1}$
