Compute $\sum\limits_n(-1)^{n-1}\frac{2^{n+1}}{2^{2n}-1}$ in terms of $\sum\limits_n\frac1{2^n-1}$ and $\sum\limits_n\frac1{2^n+1}$ Question: 

let  $$\sum_{n=1}^{+\infty}\dfrac{1}{2^n-1}=E,\sum_{n=1}^{+\infty}\dfrac{1}{2^n+1}=F$$ where $E,F$ are constant,(in fact,$E$ is  Erdős-Borwein Constant ),Find the sum
  $$f=\sum_{n=1}^{+\infty}(-1)^{n-1}\dfrac{2^{n+1}}{2^{2n}-1}$$

I tried this which I found in my texbook which seems relative, but Im not sure how to apply it to the problem:
$$f=\sum_{n=1}^{+\infty}(-1)^{n-1}\left(\dfrac{1}{2^n+1}+\dfrac{1}{2^n-1}\right)=\sum_{n=1}^{+\infty}\dfrac{(-1)^{n-1}}{2^n-1}+\sum_{n=1}^{+\infty}\dfrac{(-1)^{n-1}}{2^n+1}$$ following can't try

It's not possible to express $f$ only with $E$ and $F$ because of 
$\displaystyle\sum_{n=1}^\infty (-1)^{n-1}\frac{2^{n+1}}{2^{2n}-1}=$
$\displaystyle -\frac{1}{2}\sum_{n=1}^\infty\frac{1}{2^n-1} +\frac{1}{2}\sum_{n=1}^\infty\frac{1}{2^n+1}+\frac{i}{2}\sum_{n=1}^\infty\frac{1}{2^n+i} -\frac{i}{2}\sum_{n=1}^\infty\frac{1}{2^n-i}$
$\displaystyle -\frac{1}{\sqrt{2}}\sum_{n=1}^\infty\frac{1}{2^n+\sqrt{2}} +\frac{1}{\sqrt{2}}\sum_{n=1}^\infty\frac{1}{2^n-\sqrt{2}}+\frac{i}{\sqrt{2}}\sum_{n=1}^\infty\frac{1}{2^n+i\sqrt{2}} -\frac{i}{\sqrt{2}}\sum_{n=1}^\infty\frac{1}{2^n-i\sqrt{2}}$ .

You can define $\enspace\displaystyle g(x):=\prod_{n=1}^\infty (1+\frac{x}{2^n})\enspace$ (with the hope, that it exists a closed form for this product) and express $\enspace f\enspace$ by $\enspace\displaystyle \frac{d}{dx}\ln g(x)\enspace$ with $\enspace x\in\{-1;1;-i;i;-\sqrt{2};\sqrt{2};-i\sqrt{2};i\sqrt{2}\}$ .
