TITLE: Compute $\sum\limits_{n=1}^\infty\frac{b_{n-1}}{n 2^n}$ where $b_n=\sum\limits_{k=0}^n\frac{2^k}{k+1}$
QUESTION [6 upvotes]: 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!

REPLY [2 votes]: This sum is the Euler transform of the series $$1-0+\frac1{3^2}-0+\frac1{5^2}-0+\frac1{7^2}-\ldots = \frac{\pi^2}{6}-\frac{\pi^2}{24}=\frac{\pi^2}{8}.$$