Next step on 'Sum of Gaps'

The sum we intend to evaluate is $\sum^{\infty}_{m=1}\left(\sum^m_{n=1}\left(\sum^n_{k=1}\frac{(-1)^{k+1}}{k}-\ln 2\right)-b_2\right) - (*).$ From this problem , we know that $\left(\sum^m_{n=1}\left(\sum^n_{k=1}\frac{(-1)^{k+1}}{k}-\ln 2\right)\right)=\int^1_0\frac{x}{(1+x)^2}\ dx-\int^1_0\frac{x(-x)^m}{(1+x)^2}\ dx$ and that $b_2=\int^1_0\frac{x}{(1+x)^2}\ dx.$ Thus, the sum labelled with $(*)$ evaluates to $\begin{aligned} \sum^{\infty}_{m=1}\left(\sum^m_{n=1}\left(\sum^n_{k=1}\frac{(-1)^{k+1}}{k}-\ln 2\right)-b_2\right) &=& \sum^{\infty}_{m=1}\left(\int^1_0\frac{x}{(1+x)^2}\ dx-\int^1_0\frac{x(-x)^m}{(1+x)^2}\ dx-b_2\right)\\ &=&-\sum^{\infty}_{m=1}\int^1_0\frac{x(-x)^m}{(1+x)^2}\ dx\\ &=&-\int^1_0\sum^{\infty}_{m=1}\frac{x(-x)^m}{(1+x)^2}\ dx \quad \textrm{(by Fubini's Theorem)}\\ &=&-\int^1_0\frac{x}{(1+x)^2}\sum^{\infty}_{m=1}(-x)^m\ dx\\ &=&-\int^1_0\frac{x}{(1+x)^2}\left(\frac{-x}{1+x}\right)\ dx\\ &=&\int^1_0\frac{x^2}{(1+x)^3}\ dx\\ &=&\ln 2-\frac{5}{8}. \end{aligned}$ As such, $b_3=\ln 2-\frac{5}{8}$ and so $\lfloor 10^5b_3\rfloor=6814.$