How find this sum $\sum_{n=1}^{\infty}n\sum_{k=2^{n-1}}^{2^n-1}\frac{1}{k(2k+1)(2k+2)}$ Find the sum
$$\sum_{n=1}^{\infty}n\sum_{k=2^{n - 1}}^{2^{n}\ -\ 1}\dfrac{1}{k(2k+1)(2k+2)}$$
My try:
note
$$\dfrac{1}{k(2k+1)(2k+2)}=\dfrac{2}{(2k)(2k+1)(2k+2)}=\left(\dfrac{1}{(2k)(2k+1)}-\dfrac{1}{(2k+1)(2k+2)}\right)$$
Then I can't 

Throughout, let $H_k\equiv \sum_{j=1}^k \frac{1}{j}$ and let $\gamma$ be the Euler-Mascheroni constant.
First, note
\begin{eqnarray}
\sum_{n=1}^{\infty} n \sum_{k=2^{n-1}}^{2^n-1} \frac{1}{k(2k+1)(2k+2)}
&=& \sum_{j=0}^{\infty} \sum_{k=2^j}^{\infty} \frac{1}{k(2k+1)(2k+2)}.
\end{eqnarray}
Mathematica says
\begin{eqnarray}
\sum_{k=2^j}^{\infty} \frac{1}{k(2k+1)(2k+2)} = -\frac{1}{2}\psi(2^j) + \psi(2^j+\frac{1}{2}) - \frac{1}{2} \psi(2^j+1)
\end{eqnarray}
where $\psi$ is the digamma function. To compute this without as much help from softare, use the original poster's expression for the summand and the fact that
$$
\psi(z+1) = -\gamma + \sum_{n=1}^{\infty} \frac{z}{n(n+z)}
$$
for any $z\in\mathbb{C}$ as long as $z\ne -1,-2,-3,\ldots$.
Using explicit formulas (see Wikipedia) for $\psi(m)$ and $\psi(m+\frac{1}{2})$ where $m$ is an integer, 
\begin{eqnarray}
-\frac{1}{2}\psi(2^j) + \psi(2^j+\frac{1}{2}) - \frac{1}{2} \psi(2^j+1) &=&
-\frac{1}{2}H_{2^j-1} - \frac{1}{2} H_{2^j} - 2 \log 2 + \sum_{k=1}^{2^j} \frac{2}{2k-1} \\
&=& -\frac{1}{2}H_{2^j-1} - \frac{1}{2} H_{2^j} - 2 \log 2 + 2(H_{2^{j+1}} - \frac{1}{2} H_{2^j}) \\
&=& -\frac{1}{2}H_{2^j-1} - \frac{3}{2} H_{2^j} - 2 \log 2 + 2 H_{2^{j+1}} \\
&=& -\frac{1}{2}(H_{2^j} -\frac{1}{2^j})  - \frac{3}{2} H_{2^j} - 2 \log 2 + 2 H_{2^{j+1}} \\
&=& \frac{1}{2^{j+1}} - 2 \log 2 + 2 (H_{2^{j+1}} - H_{2^j}).
\end{eqnarray}
Observe
\begin{eqnarray}
\sum_{j=0}^m \frac{1}{2^{j+1}} - 2 \log 2 + 2 (H_{2^{j+1}} - H_{2^j})
&=& 1 - 2^{-(m+1)} + 2\left(H_{2^{m+1}}- \log 2^{m+1}\right) - 2H_1.
\end{eqnarray}
It follows
\begin{eqnarray}
\sum_{j=0}^{\infty} \sum_{k=2^j}^{\infty} \frac{1}{k(2k+1)(2k+2)} &=& \lim_{m\to\infty} 1 - 2^{-(m+1)} + 2\left(H_{2^{m+1}}- \log 2^{m+1}\right) - 2H_1 \\
&=& 2\gamma - 1.
\end{eqnarray}
