Find sum of power series. Having a small mistake. Find the sum of the series. My answer is $-\frac{3}{4}$, but it should be $\frac{3}{4}$. Where did i make a mistake?
$$ \sum_{n=1}^{\infty} \frac{n}{3^n} $$
$$ \frac{d}{dx} (\frac{1}{1-x}) = \sum_{n=0}^{\infty} \frac{d}{dx} (x^n) $$
$$ \frac{-1}{(1-x)^2} = \sum_{n=0}^{\infty} n x^{n-1} $$
$$ \frac{-x}{(1-x)^2} = \sum_{n=0}^{\infty} n x^{n} $$
Substitute $\frac{1}{3} $ into x
$$ \frac{-\frac{1}{3}}{(1-\frac{1}{3})^2} = \frac{-3}{4} $$

Since $(1-x)'=-1$,
$$
\frac{d}{dx}\frac{1}{1-x}=-\frac{(1-x)'}{(1-x)^2}=\frac{1}{(1-x)^2}
$$
by chain rule. Now the correct answer will appear.
