1. ## Evaluating Infinite Sums

Evaluate the following infinite sums. (In most cases they are f(a) where a is some obvious number and f(x) given by some power series. To evaluate the various power series, manipulate them until some well-known power series emerge.)

[sum from n=0 to inf.] n/(2^n)

Appreciate any help

2. $\displaystyle f\left( x \right) = \sum\limits_{n = 0}^\infty {x^n } = \frac{1} {{1 - x}}{\text{ }}\forall \left| x \right| < 1{\text{ }} \Rightarrow f'\left( x \right) = \sum\limits_{n = 0}^\infty {nx^{n - 1} } = \frac{1} {{\left( {1 - x} \right)^2 }}$

Now you put $\displaystyle x=\dfrac{1}{2}$ and you make some repare

3. Originally Posted by Nacho
$\displaystyle f\left( x \right) = \sum\limits_{n = 0}^\infty {x^n } = \frac{1} {{1 - x}}{\text{ }}\forall \left| x \right| < 1{\text{ }} \Rightarrow f'\left( x \right) = \sum\limits_{{\color{red}n = 1}}^{\infty} {nx^{n - 1} } = \frac{1} {{\left( {1 - x} \right)^2 }}$

Now you put $\displaystyle x=\dfrac{1}{2}$ and you make some repare
Correction.

4. Hello, blorpinbloo!

Evaluate: .$\displaystyle \sum^{\infty}_{n=0} \frac{n} {2^n}$

$\displaystyle \begin{array}{cccccc} \text{We have:} & S &=& \dfrac{1}{2} + \dfrac{2}{2^2} + \dfrac{3}{2^3} + \dfrac{4}{2^4} + \dfrac{5}{2^5} + \hdots \\ \\[-3mm] \text{Multiply by }\frac{1}{2}: & \dfrac{1}{2}S &=& \quad\;\; \dfrac{1}{2^2} + \dfrac{2}{2^3} + \dfrac{3}{2^4} + \dfrac{4}{2^5} + \hdots \end{array}$

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

The geometric series has the sum: .$\displaystyle \frac{\frac{1}{2}}{1-\frac{1}{2}} \;=\;1$

Hence, [1] becomes: .$\displaystyle \frac{1}{2}S \:=\:1$

Therefore: .$\displaystyle S \;=\;2$

5. Originally Posted by General
Correction.
Thanks, but I think that is not important, because the first terme of the sum is zero, then is same if the sum beginning from zero or one