Starting with the geometric series 1+x+x^2+x^3+... (-1<x<1), find the power series for x/[(1-x)^2] in powers of x.
Where is the series valid?
Find the sum ∑(1,∞) n/(2^n).
What have you tried already?
You are missing the fact that the geometric series $\displaystyle 1+x+x^2+x^3...=\sum_{k=0}^{\infty}x^k$ is $\displaystyle \frac{1}{1-x}$ when |x|<1
If $\displaystyle \frac{d}{dr} \sum_{k=0}^{\infty}x^k=\sum_{k=0}^{\infty}kx^{k-1}$ differentiate the left hand side from above to get the same thing in a different form. With me so far?
$\displaystyle \frac{d}{dx} \frac{1}{1-x}=\frac{1}{(1-x)^2}$
That is equal to the derivative of the power series I showed previously. So you have a power series for the above, now all you need to do is account for an x in the numerator and you will have an answer.
From my last post, the above is equal to $\displaystyle \sum_{k=0}^{\infty}kx^{k-1}$. Now multiply this by x. x represents a constant thus can be included inside the summation. Simplification leads to the power series of the expression from your problem.
here's a way on doing this without the above result:
$\displaystyle \sum\limits_{j=1}^{\infty }{\frac{j}{2^{j}}}=\sum\limits_{j=1}^{\infty }{\sum\limits_{k=1}^{j}{\frac{1}{2^{j}}}}=\sum\lim its_{k=1}^{\infty }{\sum\limits_{j=k}^{\infty }{\frac{1}{2^{j}}}}=\sum\limits_{k=1}^{\infty }{\sum\limits_{j=0}^{\infty }{\frac{1}{2^{j+k}}}}=\left( \sum\limits_{k=1}^{\infty }{\frac{1}{2^{k}}} \right)\left( \sum\limits_{j=0}^{\infty }{\frac{1}{2^{j}}} \right),$
hence your sum equals
$\displaystyle \frac{\frac{1}{2}}{1-\frac{1}{2}}\cdot \frac{1}{1-\frac{1}{2}}=2.$