Prove $\sum_{n=1}^{\infty}\frac{n-1}{2^{n+1}}=\frac{1}{2}$, without using any $\sum\frac{n}{2^{n}}$ Problem: Prove $\sum_{n=1}^{\infty}\frac{n-1}{2^{n+1}}=\frac{1}{2}$, without using any $\sum\frac{n}{2^{n}}$.
I manage to make progress, but then end up always getting a $\sum\frac{n}{2^{n}}$. For example, $\sum_{n=1}^{\infty}\frac{n-1}{2^{n+1}}$ = $\sum_{n=0}^{\infty}\frac{n}{2^{n+2}}$ = $\frac{1}{4}\sum_{n=0}^{\infty}\frac{n}{2^{n}}$, which is the sum we're not allowed to use. 
Any help on how to proceed would be appreciated!
 A: Here is a fairly general approach.  Note that $n-1=\sum_{m=1}^{n-1}(1)$ so that 
$$\begin{align}
\sum_{n=1}^\infty (n-1)a_n &=\sum_{n=2}^\infty (n-1)a_n\\\\
&=\sum_{n=2}^\infty \sum_{m=1}^{n-1} a_n\\\\
&=\sum_{m=1}^\infty \sum_{n=m+1}^\infty a_n\tag1
\end{align}$$
Now, take $a_n=x^n$, $|x|<1$ in $(1)$ and using $\sum_{n=m+1}^\infty x^n=\frac{x^{m+1}}{1-x}$ we have
$$\begin{align}
\sum_{n=1}^\infty (n-1)x^n&=\sum_{m=1}^\infty \frac{x^{m+1}}{1-x}\\\\
&=\frac{x^2}{(1-x)^2}\tag2
\end{align}$$
Using $x=1/2$ in $(2)$ yields
$$\sum_{n=1}^\infty \frac{n-1}{2^n}=1$$
from which we see that 
$$\begin{align}
\sum_{n=1}^\infty \frac{n-1}{2^{n+1}}&=\frac12
\end{align}$$
A: Render
$S=\dfrac{0}{2^1}+\dfrac{1}{2^2}+\dfrac{2}{2^3}+...$
Divide by $2$:
$S/2=\dfrac{0}{2^2}+\dfrac{1}{2^3}+\dfrac{2}{2^4}+...$
Now take the difference between terms with like denominators, e.g. $1/2^2$ from the first equation minus $0/2^2$ from the second.  This gives
$S-S/2=\dfrac{0}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...$
where all nonzero terms on the right side have $1$ as numerator.  The left side is $S/2$ and the right side may be summed as a geometric series, from which we then extract $S$.
This is essentially the way to prove the sum of a series where an arithmetic sequence is multiplied by a geometric one.
A: Let $ n $ be an integer greater than $ 1 $, observe that : $$ \frac{n}{2^{n}}-\frac{n+1}{2^{n+1}}=\frac{n-1}{2^{n+1}} $$
Which means it is a telescopic series.
