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?
 A: This just begs to be telescoped:
$$\sum_{k=2}^\infty\sum_{n=1}^k \frac{1}{k(k-1)n} = \sum_{k=2}^\infty \frac{1}{k(k-1)} + \sum_{n=2}^\infty \frac{1}{n} \sum_{k=n}^\infty\frac{1}{k(k-1)}$$
$$=  \sum_{k=2}^\infty (\frac{1}{k-1} -\frac{1}{k}) + \sum_{n=2}^\infty \frac{1}{n} \sum_{k=n}^\infty(\frac{1}{k-1} -\frac{1}{k})$$
$$= 1 + \sum_{n=2}^\infty \frac{1}{n}\cdot\frac{1}{n-1} =1 + \sum_{n=2}^\infty(\frac{1}{n-1}-\frac{1}{n}) =2$$
A: 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 $H_n/(n(n-1) \le 1/(n-1) \to 0.$
A: $$\sum_{k=2}^{\infty} \frac{H_k}{k(k-1)}=\sum_{k=1}^{\infty} \frac{H_{k+1}}{k(k+1)}=-\int_0^1 \ln(1-x)\sum_{k=1}^\infty \frac{x^k}{k}dx$$
$$=\int_0^1\ln^2(1-x)dx=\int_0^1 \ln^2xdx=2$$
Note that we use $\int_0^1 x^k \ln(1-x)dx=-\frac{H_{k+1}}{k+1}$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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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}
