# Finding the sum of a series

• Nov 1st 2007, 08:29 PM
SnowWhite
Finding the sum of a series
Hi. I need help showing the work for the sum of the series (N)/(2^N) from N=1 to infinity. The answer is two, but I'm not quite sure how to go about it.
• Nov 2nd 2007, 02:58 AM
Plato
$\displaystyle \sum\limits_{k = 0}^\infty {x^k } = \frac{1}{{1 - x}},\;\left| x \right| < 1$
$\displaystyle \sum\limits_{k = 1}^\infty {kx^{k - 1} } = \frac{1}{{^{\left( {1 - x} \right)^2 } }},\; \Rightarrow \sum\limits_{k = 1}^\infty {kx^k } = \frac{x}{{^{\left( {1 - x} \right)^2 } }}$

Let x = $\displaystyle \frac{1}{2}$.
• Nov 2nd 2007, 05:35 AM
Soroban
Hello, SnowWhite!

Another approach . . .

Quote:

Find: .$\displaystyle \sum^{\infty}_{n=1}\frac{n}{2^n}$

We have: . . . . .$\displaystyle S \;\;=\;\;\frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \frac{4}{2^4} + \frac{5}{2^5} + \cdots$

Multiply by $\displaystyle \frac{1}{2}\!:\;\;\frac{1}{2}S \;\;=\;\;\qquad\frac{1}{2^2} + \frac{2}{2^3} + \frac{3}{2^4} + \frac{4}{2^5} + \cdots$

Subtract: . . . . $\displaystyle \frac{1}{2}S \;\;= \;\;\underbrace{\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \frac{1}{2^5} + \cdots}_{\text{geometric series, sum 1}}$

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

• Nov 2nd 2007, 08:26 AM
SnowWhite
Thank you!
Thank you both so much!