# From \frac{1}{1 -x }= \sum_{n=0}^{\infty } x^n, find the sum of the series \sum_{n=1}^{\infty }...

## Question:

From {eq}\frac{1}{1 -x }= \sum_{n=0}^{\infty } x^n{/eq}, find the sum of the series {eq}\sum_{n=1}^{\infty } n(\frac{1}{3})^{n -1}{/eq}.

## Power Series:

Any polynomial can easily be expressed as a power series.

For example, The power series of {eq}\displaystyle \sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdot\cdot\cdot\\ \displaystyle \frac{1}{1-ax}=1+ax+(ax)^2+(ax)^3+(ax)^4+\cdot\cdot\cdot\\ \displaystyle \frac{1}{(1-ax)^2}=1+2ax+3(ax)^2+4(ax)^3+5(ax)^4+\cdot\cdot\cdot\\ \displaystyle e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\cdot\cdot\cdot\\ \displaystyle \ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+\cdot\cdot\cdot\\ \displaystyle \ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\frac{x^5}{5}-\cdot\cdot\cdot\\ {/eq}

Some Important Formulae:

{eq}\displaystyle \frac{d(x^n)}{dx}=nx^{n-1} {/eq}

The given power series is {eq}\displaystyle \frac{1}{1 -x }= \sum_{n=0}^{\infty } x^n{/eq}

Now, On differentiating the given series we get {eq}\displaystyle \frac{d(\frac{1}{1 -x })}{dx}= \sum_{n=0}^{\infty } \frac{d(x^n))}{dx}\\ \displaystyle \frac{-1}{(1-x)^2}=\sum_{n=0}^{\infty } nx^{n-1} {/eq}

Therefore, The sum of the series of {eq}\displaystyle \sum_{n=1}^{\infty } n(\frac{1}{3})^{n -1}=\frac{-1}{(1-\frac{1}{3})^2}\\ \displaystyle \sum_{n=1}^{\infty } n(\frac{1}{3})^{n -1}=\frac{-9}{2}\\ {/eq}