Sum 1

This is a solution that requires no knowledge of special functions.

Define the function $F(x)$ : $F(x):=\sum_{n=1}^\infty {\left(\frac{1}{n} {\left(-\frac{1}{2}\right)}^n b_{n-1} {\left(x-1\right)}^n\right)}$ Note that $F(1)=0$ and $F(0)$ is the answer to the problem.

The presence of the $\dfrac{1}{n} {\left(x-1\right)}^n$ term inspires us to take the derivative of $F(x)$ : $\begin{aligned} F'(x) &=\sum_{n=1}^\infty {\left({\left(-\frac{1}{2}\right)}^n b_{n-1} {\left(x-1\right)}^{n-1}\right)} \\ &=\sum_{n=0}^\infty {\left({\left(-\frac{1}{2}\right)}^{n+1} b_n {\left(x-1\right)}^n\right)} \\ &=-\frac{1}{2} \sum_{n=0}^\infty {\left(\frac{{\left(-1\right)}^n}{2^n} b_n {\left(x-1\right)}^n\right)} \\ &=-\frac{1}{2} \sum_{n=0}^\infty {\left(\frac{{\left(-1\right)}^n}{2^n} \sum_{k=0}^n {\left(\frac{2^k}{k+1}\right)} {\left(x-1\right)}^n\right)} \\ &=-\frac{1}{2} \sum_{n=0}^\infty {\sum_{k=0}^n {\left(\frac{{\left(-1\right)}^n}{2^{n-k}\left(k+1\right)}{\left(x-1\right)}^n\right)}} \\ &=-\frac{1}{2} \sum_{n=0}^\infty {\sum_{k=0}^n {\left(\frac{{\left(-1\right)}^k {\left(-1\right)}^{n-k}}{2^{n-k}\left(k+1\right)}{\left(x-1\right)}^k{\left(x-1\right)}^{n-k}\right)}} \\ &=-\frac{1}{2} \sum_{n=0}^\infty {\sum_{k=0}^n {\left(\frac{{\left(-1\right)}^k}{k+1}{\left(x-1\right)}^k \cdot \frac{{\left(-1\right)}^{n-k}}{2^{n-k}}{\left(x-1\right)}^{n-k}\right)}} \\ &=-\frac{1}{2} \sum_{k=0}^\infty {\sum_{n=k}^\infty {\left(\frac{{\left(-1\right)}^k}{k+1}{\left(x-1\right)}^k \cdot \frac{{\left(-1\right)}^{n-k}}{2^{n-k}}{\left(x-1\right)}^{n-k}\right)}} \\ &=-\frac{1}{2} \sum_{k=0}^\infty {\sum_{n=0}^\infty {\left(\frac{{\left(-1\right)}^k}{k+1}{\left(x-1\right)}^k \cdot \frac{{\left(-1\right)}^n}{2^n}{\left(x-1\right)}^n\right)}} \\ &=-\frac{1}{2} \sum_{k=0}^\infty {\left(\frac{{\left(-1\right)}^k}{k+1}{\left(x-1\right)}^k\right)} \sum_{n=0}^\infty {\left(\frac{{\left(-1\right)}^n}{2^n}{\left(x-1\right)}^n\right)} \\ &=-\frac{1}{2} \left(\frac{\ln{\left(1+\left(x-1\right)\right)}}{x-1}\right) \left(\frac{1}{1-\left(-\frac{x-1}{2}\right)}\right) \\ &=-\frac{\ln{x}}{x^2-1} \end{aligned}$ From here, it is possible to use the dilogarithm function to finish off the problem, but to solve it without knowledge of that function, we can recognize $-\dfrac{\ln{x}}{x^2-1}$ in integral form: $F'(x) = -\frac{\ln{x}}{x^2-1} = -\int_0^\infty {\left(\frac{y}{\left(1+y^2\right)\left(1+x^2y^2\right)}\,dy\right)}$ Integrating both sides with respect to $x$ , including the constant of integration $C$ : $F(x)=-\int_0^\infty {\left(\frac{\tan^{-1}{\left(xy\right)}}{1+y^2}\,dy\right)}+C$ Recall that $F(1)=0$ : $F(1)=-\int_0^\infty {\left(\frac{\tan^{-1}{y}}{1+y^2}\,dy\right)}+C=-{\left.\left(\frac{{\left(\tan^{-1}{y}\right)}^2}{2}\right)\right\rvert}_0^\infty+C=-\left(\left(\frac{{\left(\frac{\pi}{2}\right)}^2}{2}\right)-\left(\frac{0^2}{2}\right)\right)+C=-\frac{\pi^2}{8}+C$ Thus, $C = \dfrac{\pi^2}{8}$ , so: $F(x)=\frac{\pi^2}{8}-\int_0^\infty {\left(\frac{\tan^{-1}{\left(xy\right)}}{1+y^2}\,dy\right)}$ and we have: $F(0)=\frac{\pi^2}{8}-\int_0^\infty {\left(\frac{0}{1+y^2}\,dy\right)}=\frac{\pi^2}{8}$ In summary, we have found that the sum is: $\sum_{n=1}^\infty {\left(\frac{b_{n-1}}{n\cdot2^n}\right)}=F(0)=\boxed{\dfrac{\pi^2}{8}}$ and the answer is $2+8=10$ .

define $B (x)= \dfrac{1}{2} \sum_{k=1}^{n+1} \dfrac{x^k}{k} \to B'(x)=\dfrac{1}{2}\sum_{k=1}^{n+1} x^{k-1}=\dfrac{x^{n+1}-1}{2(x-1)}\\ B(2)= \dfrac{1}{2} \int_0^2 \dfrac{x^{n+1}-1}{x-1} dx$ note $B(2)$ is the partial sum asked in the question,i.e, $b_n$ . the summation asked is then simply $\dfrac{1}{2}\sum_{n=1}^{\infty} \dfrac{1}{n.2^n}\int_0^2 \dfrac{x^{n}-1}{x-1} dx=\dfrac{1}{2}\int_0^2\dfrac{1}{x-1}\sum_{n=1}^{\infty} \left(\dfrac{(x/2)^n}{n}-\dfrac{1}{n.2^n}\right)dx\\=\dfrac{1}{2}\int_0^2\dfrac{-\ln(1-x/2)+\ln(1/2)}{x-1}dx=\dfrac{1}{2}\int_0^2\dfrac{\ln(2-x)}{1-x}dx$ substituting $u=x-1$ we have $\dfrac{1}{2}\int_{-1}^1\dfrac{\ln(1-u)}{-u}du=\dfrac{1}{2}(li_2(1)-li_2(-1))=\dfrac{\pi^2}{8}$