Collecting proofs for $\sum_{n=2}^{\infty} \, \frac{n-1}{2^n} = 1$ Update: The summation I came across has the form shown in title, and that exact question appears to be new. I could ask for proofs that take on this summation directly (without reducing it to summations starting at $n = 0$ or $n =1$), and that would be the preferred answer and the only one I will accept. But might as well let this fly as is, collecting all proofs of the identified equivalent variants.
I just stumbled across the fact that
$\tag 1 \sum_{n=2}^{\infty} \, \frac{n-1}{2^n} = 1$
This is equivalent to
$\tag 2 \sum_{n=1}^{\infty} \, \frac{n}{2^n} = 2$
I discovered $\text{(1)}$ using a 'matrix/combinatorial' argument, but it would need work to turn it into a formal proof.
I googled and found this Quora link, explaining how to show $\text{(2)}$.
I didn't find the question on this site, prompting this 'collecting proofs post':

Please supply a proof demonstrating either $\text{(1)}$ or
$\text{(2)}$. If you use any theory or technique, mention that at the
start of your answer.

 A: A probability theory flavored approach: The expression $\sum_{n\geq1} \frac{n}{2^n}$ is also the expected number of IID fair coin flips it takes to gets a heads. Assuming you know this number is finite (by some root/ratio test business), let $H$ be the expected time. Then from conditioning on seeing heads or tails on the first flip, respectively, $H$ satisfies the recursion
$$H = \frac{1}{2}(1) + \frac{1}{2}(H+1),$$
so $H = 2$. 

A brute-force approach: by induction or otherwise,
$$\sum_{k=1}^n \frac{k}{2^k} = 2 - \frac{n+2}{2^n}.$$
Sending $n \to \infty$ recovers the desired result.
A: One approach is to note that for $S_k:=\sum_{n\geq k}\frac{1}{2^n}$ you have $S_0=2$ and $S_{k+1}=\frac{1}{2}S_k$. Now rearranging terms in the summation yields
$$\sum_{n\geq1}\frac{n}{2^n}
=\sum_{n\geq1}\sum_{k=1}^n\frac{1}{2^n}
=\sum_{k\geq1}\sum_{n\geq k}\frac{1}{2^k},$$
corresponding to the following picture:

The inner sums equal the $S_k$, so we can simplify this to
$$\sum_{k\geq1}S_k
=\sum_{k\geq1}\frac{1}{2^k}S_0
=S_0\sum_{k\geq1}\frac{1}{2^k}=S_0S_1=2.$$
A: We have that for $|x|<1$ by
$$f(x)=\sum_{k\ge1} x^n=\frac x{1-x} \implies f'(x)=\sum_{k\ge1} nx^{n-1}=\frac1{(1-x)^2}$$
therefore
$$\sum_{k\ge1} nx^{n}=x\cdot \sum_{k\ge1} nx^{n-1}=\frac x{(1-x)^2}$$
A: The standard technique is $\sum_{n\ge 1}r^n=\frac{1}{1-r}-1\implies\sum_{n\ge 1}nr^{n-1}=\frac{1}{(1-r)^2}$.
A: A solution by examining the differences between successive terms.
Let us define
$$S = \sum_{n=1}^{\infty} \, \frac{n}{2^n} = \sum_{n=1}^{\infty} u_n$$
Note (Edit): the definition $u_n =  \frac{n}{2^n}$ will be used for $n=0$, i.e. for a term not involved in the series.
We have:
$$u_n - u_{n+1} = \frac{n-1}{2^{n+1}} =  \frac{u_{n-1}}{4}$$
It follows immediately:
$$0 = S - S =  \sum_{n=1}^{\infty} u_n -  \sum_{n=1}^{\infty} u_{n+1} -u_1 = -u_1  + \frac{u_0}{4} + \frac{S}{4}  $$
And therefore, noting that $u_0=0$ and $u_1 = \frac{1}{2}$
$$ S  = \sum_{n=1}^{\infty} \, \frac{n}{2^n} = 2 $$
Note: the same procedure can be used to show that
$$ \sum_{k=1}^n \frac{k}{2^k} = 2 - \frac{n+2}{2^n} $$
A: I stumbled on this by realizing that since 
$$ \tag 1 \sum_{n=1}^{\infty} \, \frac{1}{2^n} = 1$$
it must be true that
$$ \tag 2 (\sum_{n=1}^{\infty} \, \frac{1}{2^n}) \times (\sum_{n=1}^{\infty} \, \frac{1}{2^n})  = 1$$
Using the rearrangement approach (and the identity $\sum_1^n \, 1 = n$) found in Servaes' answer, 
$$ \tag 3 (\sum_{n=1}^{\infty} \, \frac{1}{2^n}) \times (\sum_{m=1}^{\infty} \, \frac{1}{2^m})  = \sum_{n+m =k}^{\infty} \, \frac{1}{2^k} = \sum_{k=2}^{\infty} \, \frac{k-1}{2^k} = 1$$
demonstrating the identity equation.
