Finding sun of $$\sum^{\infty}_{k=1}\frac{(-1)^{k-1}}{k}\sum^{\infty}_{r=0}\frac{1}{k2^{r}+1}$$
what i try
Let $$S=\sum^{\infty}_{k=1}\frac{(-1)^{k-1}}{k}\sum^{\infty}_{r=0}\frac{1}{k2^{r}+1}$$
$$S=\sum^{\infty}_{k=1}\frac{(-1)^{k-1}}{k}\bigg(\frac{1}{k+1}+\frac{1}{2k+1}+\frac{1}{4k+1}+\cdots\bigg)$$
How do i solve it Help me please
With CAS help:
$$\sum _{k=1}^{\infty } \sum _{r=0}^{\infty } \frac{(-1)^{k-1}}{\left(k 2^r+1\right) k}=\\\sum _{k=1}^{\infty } \sum _{r=0}^{\infty } \mathcal{M}_a\left[\frac{(-1)^{k-1}}{\left(k 2^r+a\right) k}\right](s)=\\\mathcal{M}_s^{-1}\left[\sum _{k=1}^{\infty } \sum _{r=0}^{\infty } (-1)^{-1+k} 2^{r (-1+s)} k^{-2+s} \pi \csc (\pi s)\right](1)=\\\mathcal{M}_s^{-1}\left[\sum _{r=0}^{\infty } \left(\sum _{k=1}^{\infty } (-1)^{-1+k} 2^{r (-1+s)} k^{-2+s} \pi \csc (\pi s)\right)\right](1)=\\\mathcal{M}_s^{-1}\left[\sum _{r=0}^{\infty } -2^{-1+r (-1+s)} \left(-2+2^s\right) \pi \csc (\pi s) \zeta (2-s)\right](1)=\\\mathcal{M}_s^{-1}[\pi \csc (\pi s) \zeta (2-s)](1)=\\\mathcal{M}_s^{-1}\left[\pi \csc (\pi s) \sum _{j=1}^{\infty } \frac{1}{j^{2-s}}\right](1)=\\\sum _{j=1}^{\infty } \mathcal{M}_s^{-1}\left[\frac{\pi \csc (\pi s)}{j^{2-s}}\right](1)=\sum _{j=1}^{\infty } \frac{\pi }{j \pi +j^2 \pi }=\\\sum _{j=1}^{\infty } \frac{1}{j (1+j)}=1$$
where: $\mathcal{M}_a[f(a)](s)$,$\mathcal{M}_s^{-1}[f(s)](a)$ is Mellin transform and Inverse Mellin transform.
