Show $ f(x)=\frac{x^2}{(1-x^2)^2}-\frac{3x^2}{(3-x^2)^2}, |x|<1 $ is equal to $ f(x)=\sum_{n=0}^\infty{n(1-3^{-n})x^{2n}}, |x|<R $

Question

Let the function $f$ be a sum-function on the interval $]-1,1[$, where $R$ is the radius of convergence. $$ f(x)=\sum_{n=0}^\infty{n(1-3^{-n})x^{2n}}, |x|<1 $$ I find that the radius of convergence was $R=1$. Now I have to find that the sum-function is equal to the following: $$ f(x)=\frac{x^2}{(1-x^2)^2}-\frac{3x^2}{(3-x^2)^2}, |x|<1 $$ I know that $$ \sum_{n=0}^\infty{r^n=\frac{1}{1-r}},|r|<1 $$ and $$ \sum_{n=0}^\infty{ar^n=\frac{a}{1-r}} $$ So I got from the following $$ f(x)=\frac{x^2}{(1-x^2)^2}-\frac{3x^2}{(3-x^2)^2} $$ to this $$ f(x)=\left(\sum_{n=0}^\infty{x^{2n+1}}\right)^2-\left(\sum_{n=0}^\infty{x^{2n+1}\cdot 3^{-n}}\right)^2 $$ I don't know what I have to do now and if I have done it right.

Thanks in advance

Answer 1

$$\sum_{n=0}^{\infty} z^{n}=\frac{1}{1-z}, |z|<1~~~(1)$$ D. w.t.$z$ both sides $$\sum_{k=0}^{\infty} nz^n =\frac{z}{(1-z)^2}~~~~(2)$$ Now $$g(t)=\frac{t}{(1-t)^2}-\frac{t/3}{(1-t/3)^2}$$ Using (2), we get $$\implies g(t)=\sum_{n=0}^{\infty}n [t^n-(t/3)^n]$$ So $$f(x)=g(x^2)=\sum_{n=0}^{\infty} n[x^{2n}-\frac{x^{2n}}{3^n}]=\sum_{n=0}^{\infty} n(1-3^{-n})x^{2n},~~ x^2<1.$$

OP needs to use (2) rather than (1). The domain of convergence (validity) of these reult is $(-1,1).$

Answer 2

First, let's start with an identity: $$ \begin{align} \sum_{n=0}^\infty nx^n = x\cdot\sum_{n=0}^\infty nx^{n-1} &= x \cdot \frac d{dx} \int_0^x \sum_{n=0}^\infty nt^{n-1} dt\\ &= x \cdot \frac d{dx} \sum_{n=0}^\infty x^n\\ &= x \cdot \frac d{dx} \frac1{1-x}\\ &= \frac x {(1-x)^2} \end{align} $$ So now, split the original sum into two: $$ \begin{align} f(x) &= \sum_{n=0}^\infty{nx^{2n}}-\sum_{n=0}^\infty{n\cdot3^{-n}x^{2n}}\\ &= \sum_{n=0}^\infty{n\left(x^2\right)^n}-\sum_{n=0}^\infty{n\left(\frac {x^2}3\right)^n}\\ &= \frac{x^2}{\left(1-x^2\right)^2}-\frac{x^2/3}{\left(1-x^2/3\right)^2}\\ &= \frac{x^2}{\left(1-x^2\right)^2}-\frac{3x^2}{\left(3-x^2\right)^2} \end{align} $$ Remember the method of using derivatives and integrals to cancel the $n$ that we used for the identity! It's very useful, and is the key to a lot of problems like this one.

Answer 3

\begin{align*} f(x)&=\frac{x^2}{(1-x^2)^2}-\frac{3x^2}{(3-x^2)^2}\\ &=\frac{x}{2}(\frac{1}{1-x^2})^{'}-\frac{3x}{2}(\frac{1}{3-x^2})^{'}\\ &=\frac{x}{2}(\sum_{n=0}^{\infty}x^{2n})^{'}-\frac{x}{2}(\sum_{n=0}^{\infty}(\frac{x}{\sqrt{3}})^{2n})^{'}\\ &=\frac{x}{2}(\sum_{n=0}^{\infty}2nx^{2n-1}-\sum_{n=0}^{\infty}3^{-n}2nx^{2n-1})\\ &=\sum_{n=0}^{\infty}nx^{2n}(1-3^{-n})\\ \end{align*}