We're given the following problem:
"We know that $\frac{1}{1 - x} = \sum_{k=0}^{\infty} x^k $ for $ -1 < x < 1 $. Using the derivative with respect to $x$, calculate the sum of the following power-series: $ \sum_{k=1}^{\infty} kx^k $ and $ \sum_{k=1}^{\infty} k^2x^k $."
I have thus first taken the derivative of the left side:
$$ \frac{d}{dx}\left[\frac{1}{1 - x}\right] = \frac{1}{(1 - x)^2}$$
And of the right side:
$$ \frac{d}{dx}\left[\sum_{k=0}^{\infty}x^k\right] = \frac{d}{dx}\left[1 + x + x^2 + x^3 + x^4 + ...\right] $$
$$ = 0 + 1 + 2x + 3x^2 + 4x^3 + ...$$
$$ = \sum_{k=0}^{\infty} kx^{k-1} $$
Then I substituted these derivatives in the following equation:
$$ \sum_{k=0}^{\infty} kx^{k-1} = \frac{1}{(1 - x)^2} $$
From this I rewrote the power-series I needed to find the sum from:
$$ \sum_{k=0}^{\infty} kx^k = \sum_{k=0}^{\infty} kx^1x^{k-1} $$ $$ = \sum_{k=0}^{\infty} x\left(\frac{1}{(1 - x)^2}\right) $$ $$ = \sum_{k=0}^{\infty} \left(\frac{x}{(1 - x)^2}\right) $$
Now I am a little stuck. The solution says the sum of the first power-series should equal $\frac{x}{(1 - x)^2}$. I think that I'm close but I don't know how to continue.
Can anyone help me out?
Yes, you are close: since$$\sum_{k=0}^\infty kx^{k-1}=\frac1{(1-x)^2},$$multiplying both sides by $x$ gives you$$\sum_{k=0}^\infty kx^k=\frac x{(1-x)^2}.$$