<s>[INST] 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!
 [/INST] 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.
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