Compute $\sum\limits_{n=1}^\infty\frac{b_{n-1}}{n 2^n}$ where $b_n=\sum\limits_{k=0}^n\frac{2^k}{k+1}$ Define the sequence $a_n := \frac{2^n}{n+1}$. Then, define the sequence $b_n$, the partial sums of $a_n$; i.e.:
$$b_n=\sum_{k=0}^{n} {a_k}$$
The problem is to compute:
$$\sum_{n=1}^{\infty} {\left(\frac{b_{n-1}}{n 2^n}\right)}$$
This is very difficult to estimate using a computer algebra system. On Maple, I cannot complete an estimate of $100000$ terms under a minute. (However, the sum of the first $1000$ terms is approximately $1.23270055$.)
I am wondering whether the double sum can be converted into a series convolution of some sort; the possibility of getting a $2^{n-m}$ term seems to help that prospect.
Thanks!
 A: Consider the following.
Given:
\begin{align}
b_{n} &= \sum_{k=0}^{n} \frac{2^{k}}{k+1} \\
S &= \sum_{n=1}^{\infty} \frac{b_{n-1}}{2^{n} \, n}.
\end{align}
Now:
$$b_{n} = \frac{1}{2} \, \sum_{k=1}^{n+1} \frac{2^{k}}{k} $$
which leads to
\begin{align}
S &= \frac{1}{2} \, \sum_{n=1}^{\infty} \sum_{k=1}^{n} \frac{2^{k}}{2^{n} \, n \, k} = \frac{1}{2} \, \sum_{n,k=1}^{\infty} \frac{1}{2^{n} \, k \, (n+k)}  = \frac{\pi^{2}}{8}.
\end{align}
An alternate view is:
\begin{align}
b_{n} &= \sum_{k=0}^{n} \frac{2^{k}}{k+1} = \sum_{k=0}^{n} 2^{k} \, \int_{0}^{1} t^{k-1} \, dt \\
&= \int_{0}^{1} \left( \sum_{k=0}^{n} (2 t)^{k} \right) \, dt \\
&= \int_{0}^{1} \frac{1 - (2 t)^{n+1}}{1-2 t} \, dt.
\end{align}
Now,
\begin{align}
S &= \sum_{n=1}^{\infty} \frac{1}{2^{n} \, n} \, \int_{0}^{1} \frac{1 - (2 t)^{n}}{1-2 t} \, dt \\
&= \int_{0}^{1} \frac{\ln(2) + \ln(1-t)}{1-2 t} \, dt \\
&= \left[ \frac{1}{2} \, Li_{2}(2 t -1) \right]_{0}^{1} = \frac{1}{2} \, (Li_{2}(1) - Li_{2}(-1) ) \\
&= \frac{1}{2} \, \left(\frac{\pi^2}{6} + \frac{\pi^2}{12} \right) = \frac{\pi^{2}}{8}. 
\end{align}
Note: $Li_{2}(x)$ is the dilogarithm function.
A: This sum is the Euler transform of the series $$1-0+\frac1{3^2}-0+\frac1{5^2}-0+\frac1{7^2}-\ldots = \frac{\pi^2}{6}-\frac{\pi^2}{24}=\frac{\pi^2}{8}.$$
