1. ## Evaluate the series

1. sum of (n/2^n) from n=1 to infinite
2. sum of (n(n-1)x^n) from n=2 to infinite

2. Hello, Jojo123!

Here's the first one . . .

$\displaystyle 1)\;\;\sum^{\infty}_{n=1} \frac{n}{2^n}$

We are given: . $\displaystyle S \;=\;\frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \frac{4}{2^4} + \frac{5}{2^5} +\hdots$

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

$\displaystyle \text{Subtract: }\qquad\quad\frac{1}{2}S \;=\;\underbrace{\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \hdots}_{\text{geometric series}}$

The geometric series has a sum of $\displaystyle 1.$

. . So we have: .$\displaystyle \frac{1}{2}S \:=\:1 \quad\Rightarrow\quad\boxed{ S \:=\:2}$

3. Originally Posted by Jojo123
2. sum of (n(n-1)x^n) from n=2 to infinite
$\displaystyle f(x)=\sum_{n=0}^\infty x^n=\frac{1}{1-x}$ (assuming |x|<1)

hence $\displaystyle f'(x)=\sum_{n=1}^\infty nx^{n-1}=\left(\frac{1}{1-x}\right)'=\frac{1}{(1-x)^2}$

$\displaystyle f''(x)=\sum_{n=2}^\infty n(n-1)x^{n-2}=\left(\frac{1}{(1-x)^2}\right)'=\frac{1}{2(1-x)^3}$

4. Originally Posted by Jojo123

2. sum of (n(n-1)x^n) from n=2 to infinite
Put $\displaystyle \alpha=\sum\limits_{n=1}^{\infty }{nx^{n}}\implies \alpha \cdot x=\sum\limits_{n=1}^{\infty }{nx^{n+1}}.$ From here it's not hard to see that $\displaystyle \alpha =\frac{x}{(1-x)^{2}}.$ Now put $\displaystyle \beta=\sum\limits_{n=1}^{\infty }{n(n-1)x^{n}}\implies \beta \cdot x=\sum\limits_{n=1}^{\infty }{n(n-1)x^{n+1}}$ and compute $\displaystyle \beta-\beta\cdot x.$