Find sum of power series. Having a small mistake.

Question

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

$$ \frac{-\frac{1}{3}}{(1-\frac{1}{3})^2} = \frac{-3}{4} $$

Your differentiation is wrong (use the chain rule, carefully). — David, Apr 06 '16 at 06:10
${d\over dx}{1\over 1-x}=+{1\over (1-x)^2}$. Two minus signs are cancelling each other one is coming from the derivation of the denominator $1-x$ and one from the derivative of an inverse $1/u$ (Where $u=1-x$) — marwalix, Apr 06 '16 at 06:14

score 1 · Accepted Answer · answered Apr 06 '16 at 06:10

1

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.

answered Apr 06 '16 at 06:10

choco_addicted

1 Answers1