Sum 1 follow-up

write the sum as $\sum_{n=1}^\infty \dfrac{(-1)^n}{n} \int_0^{-1}\dfrac{x^n-1}{x-1} dx=\int_0^{-1} \dfrac{\ln(1+x)-\ln(2)}{x-1} dx$ substituting $x=1-2u$ we have $\int_{1/2}^1 \dfrac{-\ln(1-u)}{u} du=li_2(1)-li_2(1/2)= \dfrac{\pi^2}{12}+\dfrac{\ln^2(2)}{2}$

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

Taking the derivative, we have: $F'(x)=\sum_{n=0}^\infty {\left(\sum_{k=1}^{n+1} {\left(\frac{\left(-1\right)^k}{k}\right)} {\left(x-1\right)}^n\right)}$ and through a very similar series of steps to that in my solution to "Sum 1" , we can simplify the above expression to: $F'(x)=\frac{\ln{x}}{\left(x-1\right)\left(x-2\right)}$ Thus (since $F(1)=0$ ): $F(x)=F(1)+\int_1^x {\left(\frac{\ln{t}}{\left(t-1\right)\left(t-2\right)}\,dt\right)}=\int_1^x {\left(\frac{\ln{t}}{\left(t-1\right)\left(t-2\right)}\,dt\right)}$ and the answer to the problem is $F(0)$ : $F(0)=\int_1^0 {\left(\frac{\ln{t}}{\left(t-1\right)\left(t-2\right)}\,dt\right)}=-\int_0^1 {\left(\frac{\ln{t}}{\left(t-1\right)\left(t-2\right)}\,dt\right)}$ In the solution to the original problem, I alluded to the possibility of solving it using the dilogarithm function ; here, the dilogarithm function is necessary: $\int_0^1 {\left(\frac{\ln{t}}{\left(t-1\right)\left(t-2\right)}\,dt\right)}=\textrm{dilog}\left(\frac{1}{2}\right)-\frac{\pi^2}{6}$ However, $\textrm{dilog}\left(\dfrac{1}{2}\right)$ is a special value of the dilogarithm function: $\textrm{dilog}\left(\frac{1}{2}\right)=\frac{\pi^2}{12}-\frac{\ln^2{2}}{2}$ In summary, we have shown that: $\begin{aligned} \sum_{n=1}^{\infty} {\left(\frac{\left(-1\right)^n}{n}\sum_{k=1}^n {\left(\frac{\left(-1\right)^k}{k}\right)}\right)} &=F(0) \\ &=-\int_0^1 {\left(\frac{\ln{t}}{\left(t-1\right)\left(t-2\right)}\,dt\right)} \\ &=-\left(\textrm{dilog}\left(\frac{1}{2}\right)-\frac{\pi^2}{6}\right) \\ &=-\left(\left(\frac{\pi^2}{12}-\frac{\ln^2{2}}{2}\right)-\frac{\pi^2}{6}\right) \\ &=\boxed{\dfrac{\pi^2}{12}+\dfrac{\ln^2{2}}{2}} \end{aligned}$ and the answer is $10000\cdot2+1000\cdot2+100\cdot2+10\cdot2+2=22222$ .