How to find the partial sum of $n/2^n$? I'm trying to find a formula for the partial sum of $n/2^n$.
I've tried this so far...
$$ S_n = \frac12 + \frac24 + \frac38 + \cdots + \frac{n}{2^n} $$
Then I tried to find a way to eliminate most of the terms by multiplying the whole sequence by $\frac{2^n}n\cdot\frac{n+1}{2^{n+1}}$ (a terms that should take each term in the sequence to the next term.)
$$ \frac{n+1}{2n} S_n =  \frac24 + \frac38 + \cdots + \frac{n+1}{2^{n+1}} $$
Then
$$ S_n- \frac{n+1}{2n} S_n = \frac12 - \frac{n+1}{2^{n+1}}$$
$$ S_n = \frac{\frac12 - \frac{n+1}{2^{n+1}}}{1-\frac{n+1}{2n}} $$
But, alas, this does not give the correct answers.
Can someone point out where I went wrong?  Thanks

Observe
$$2S_n = 1+ \frac{1+1}2 + \frac{2+1}4 + \cdots + \frac{(n-1)+1}{2^{n-1}}
\\=S_{n-1}+1+\frac12+\frac14\cdots+\frac1{2^{n-1}}
\\=S_n-\frac n{2^n}+2-\frac1{2^{n-1}}.$$
