Summation of simple sequence

Question

This is a fairly simple question, I'm sure, but I appear to be having trouble. What is the result of the following sequence:

$$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+ .... + \frac{n}{2^{n}}.$$ ?

Thanks

Answer 1

$$2S_n-S_n=\left(\frac11+\frac{2}{2}+\frac{3}{4}+\cdots+ \frac{n}{2^{n-1}}\right)-\left(\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+ \cdots+ \frac{n}{2^{n}}\right)\\ =1+\frac{1}{2}+\frac{1}{4}+ \cdots+ \frac{1}{2^{n-1}}-\frac n{2^n}=2-\frac{n+2}{2^n}.$$

Answer 2

Hint: Taking the derivative of $\sum_{i=0}^n p^i=\frac{1-p^{n+1}}{1-p}$ with respect to $p$ and some manipulation will give you the result.

Answer 3

$$\begin{align} \sum_{k=1}^n\frac{k}{2^k}&=\sum_{k=1}^n\frac{1}{2^k}\sum_{j=1}^k(1)\\\\ &=\sum_{j=1}^{n}\sum_{k=j}^{n}\frac{1}{2^k}\\\\ &=\sum_{j=1}^{n}\frac{(1/2)^j-(1/2)^{n+1}}{1-(1/2)}\\\\ &=2\sum_{j=1}^{n}\left((1/2)^j-(1/2)^{n+1}\right)\\\\ &=-2n(1/2)^{n+1}+2\frac{(1/2)-(1/2)^{n+1}}{1-(1/2)}\\\\ &=2-(1/2)^{n-1}-n(1/2)^n \end{align}$$