Re: Sum of a given series

You can prove that the sum converges to 2 by considering the partial sums.

It turns out that (I encourage you to prove this):

$\displaystyle \sum_{k=1}^n \frac{k}{2^k} = \frac{2^{n+1}-n-2}{2^n}$.

Hence, taking the limit as n tends to infinity, you can see that the sum converges to 2.

Re: Sum of a given series

Hello, Jame!

Quote:

I was wondering if there was any way to calculate the sum of the series

. . $\displaystyle \sum_{k=1}^\infty \frac{k}{2^k}$

$\displaystyle \begin{array}{ccccc}\text{We have: }& S & =& \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \frac{4}{2^4} + \cdots \\ \\[-4mm] \text{Multiply by }\frac{1}{2}\!: & \frac{1}{2}S &=& \quad\;\; \frac{1}{2^2} + \frac{2}{2^3} + \frac{3}{2^4} + \cdots \\ \\[-4mm] \text{Subtract: }& \frac{1}{2}S &=& \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \cdots \end{array}$

The right side is a geometric series

. . with first term $\displaystyle a = \tfrac{1}{2}$ and common ratio $\displaystyle r = \tfrac{1}{2}.$

Its sum is: .$\displaystyle \frac{\frac{1}{2}}{1-\frac{1}{2}} \:=\:1$

Therefore: .$\displaystyle \tfrac{1}{2}S \:=\:1 \quad\Rightarrow\quad S \:=\:2$