$\sum_{n=1}^\infty {2\beta_n}\frac{1}{(1-\beta_n)(1-\beta_n/2) }$ where $\beta=\frac{1}{2^n}$ I'm calculating an infinite series, and halfway I generate another series I can't yet calculate. (A note for myself: I expand the original series and then sum it up and get a series which is not very different from the original one.)
$\sum_{n=1}^\infty {2\beta_n}\frac{1}{(1-\beta_n)(1-\beta_n/2) }$ where $\beta=\frac{1}{2^n}$
I'm working on it, but any hint would be welcome.
 A: Hint. Observe that, with $\beta:=\frac{1}{2^n}$,
$$
{2\beta_n}\frac{1}{(1-\beta_n)(1-\beta_n/2)}=\frac{4}{1-2^{-n}}-\frac{4}{1-2^{-(n+1)}}
$$ giving a telescoping series.
A: $$S=4\sum_{n=1}^{\infty}\frac{\beta_n}{(1-\beta_n)(2-\beta_n)}= 4\sum_{n=1}^{\infty}\left(\frac{1}{1-\beta_n}-\frac{2}{2-\beta_n}\right)$$
$$\implies S=4 \sum_{n=1}^{\infty}\left(\frac{1}{1-\beta_n}-\frac{1}{1-\beta_{n+1}}\right)$$
Further by telescopic summation we get
$$S=4\left(\frac{1}{1-1/2}-\frac{1}{1-\beta_{n+1}}\right)=8\left(1-\frac{2^n}{2^{n+1}-1}\right).$$
A: My partial solution:
$\sum_{n=1}^\infty {2\beta_n}\frac{1}{(1-\beta_n)(1-\beta_n/2) }
=2 \sum_{n=1}^\infty \frac{2\beta_n}{(1-\beta_n)}-2\sum_{n=1}^\infty \frac{2\beta_n/2}{(1-\beta_n/2)}  ,$ (Can we change the order of summation here?)
Thus we break the series to two convergent series.
Or by expanding the infinite series we can get it equals
$2\sum\beta_(1+\beta_n+\beta_n^2+\dots)(1+\beta_n+\beta_n^2/2+\beta_n^3/4+\dots)\\
=2\sum \beta_n\{1+(1+1/2)\beta_n+(1+1/2+1/4)\beta_n^2+\dots\}\\
=2\times2\sum(\beta_n+\beta_n^2+\dots)-2\sum(\beta_n+\beta_n^2/2+\beta_n^3/4+\dots),$      (1)
which produces the same result. This approach, though more complicated, does reveal we can do elementary algebraic operation by Taylor expansion. I guess this way is similar to Ntn's initial  idea in calculus of expanding a function to power series, and this way we eliminate the need of 'division'. More importantly, it shows a right way to separate the original series to two convergent series, for example, we can also have $\sum {2\beta_n}\frac{1}{(1-\beta_n)(1-\beta_n/2) }=4 \{\sum\frac{1}{(1-\beta_n) }-\sum\frac{1}{(1-\beta_n/2) }\}$, but neither of the two converges.
(A further note about the above expansion: how we calculate $1+\beta_n+\beta_n^2/2+\beta_n^3/4+\dots$? Let $A=1+\beta_n+\beta_n^2+\dots, B=1+\beta_n+\beta_n^2/2+\beta_n^3/4+\dots$, then obviously from (1), $AB=2A-B,  B=2A/(1+A)=1/(1-\beta_n/2)$; or we can calculate it by $(\beta_n-1)B=-2\beta_n+B$.)

Solution 1:
The above effort is going around in circles. Inspired by the comment @ aditya gupta, I notice I need to use recursive relation, namely
let $f(n)=\frac{\beta_n}{(1-\beta_n)}$ (simply 1/(2$^n$-1)), notice $f(n+1)=\frac{\beta_n/2}{(1-\beta_n/2)}$, then $\sum_{n=1}^\infty {2\beta_n}\frac{1}{(1-\beta_n)(1-\beta_n/2) }\\
=2 \sum_{n=1}^\infty \frac{2\beta_n}{(1-\beta_n)}-2\sum_{n=1}^\infty \frac{2\beta_n/2}{(1-\beta_n/2)} \\
=4\sum_{n=1}^\infty f(n)-4\sum_{n=1}^\infty f(n+1)=4\lim_{N\to \infty} f(1)-f(N+1)=4.$
This shows how important it is to go in the right direction.

Solution 2:
This another solution is inspired by @Olivier Oloa. Note it's also almost the same as that of @Dr Zafar Ahmed DSc.
Notice though in $4 \{\sum\frac{1}{(1-\beta_n) }-\sum\frac{1}{(1-\beta_n/2) }\}$, neither converges, (using recursive relation again,) it equals
$4 \{\sum\frac{1}{(1-\beta_n) }-\frac{1}{(1-\beta_n/2) }\}
=4 \sum\{\frac{1}{(1-\beta_n) }-\frac{1}{(1-\beta_{n+1} )}\}$ which converges conditionally to $4\lim_{N\to \infty}\frac{1}{(1-\beta_1) }-\frac{1}{(1-\beta_{N+1} )} =4\times(\frac{1}{1-1/2}-\frac{1}{1-0})=4.$

In a word there are two ways of changing the series to telescope series, one is to make it absolutely convergent series $\sum$1/(2$^n$-1)-1/(2$^{n+1}$-1), another is to make it conditionally convergent series $\sum$1/(1-2$^{-n}$)-1/(1-2$^{-n-1}$).
PS: another example of telescope series is m$^z$=$\prod_n$((n+1)/n)$^z$.
