Evaluate $\sum\limits_{k=1}^{n} \frac{k}{2^k}$

Question

Evaluate $$\sum\limits_{k=1}^{n} \frac{k}{2^k}$$

Answer 1

HINT: Let $$S = \sum_{k = 1}^{n}(\frac{k}{2^k})$$

Try evaluating $S - \frac{S}{2}$ and you will see the trend.

Answer 2

You have that:

$$\displaystyle\sum_{k=0}^n r^k=\displaystyle\frac{1-r^{n+1}}{1-r}$$

derive the both side to get an expression to this sum, observe that your first term is $1$ not zero. Take $r=\displaystyle\frac{1}{2}$.

Answer 3

$$\sum\limits_{k=1}^{n} \frac{k}{2^k} = \frac{1}{2} \sum_{k=1}^{n} \frac{k}{2^{k-1}} = \frac{1}{2} \left(\sum_{k=1}^{n} \frac{k - 1}{2^{k-1}} + \sum_{k=1}^{n} \frac{1}{2^{k-1}}\right) = ....$$

Good?