show that the sum function f is given by F

Question

I have a problem that sounds.

Let f be given as the sum function on the interval ]-R,R[ $$f(x)=\sum_{n=0}^{\infty}n(1-3^{-n})x^{2n}$$ for $$|x|<R$$

Show that

$$f(x)=\frac{x^2}{(1-x^2)^2}-\frac{3x^2}{(3-x^2)^2}$$ for $$|x|<R$$

First my idea os to look at it separted $$\sum_{n=0}^{\infty}n(1-3^{-n})x^{2n}=\sum_{n=0}^{\infty}nx^{2n}-\sum_{n=0}^{\infty}(1-\frac{1}{3^n})x^{2n}$$ My idea is then to use some geometric series properties to rewrite the expression without the sum sign, but how I am unsure of?

Prior to this, I have determined the radius of convergence to $$ R=1$$

Thanks in advance

What??? $n(1-3^{-n})\ne n-(1-3^{-n})$. – David C. Ullrich Aug 07 '20 at 12:30 — David C. Ullrich, Aug 07 '20 at 12:30

score 1 · Answer 1 · answered Aug 07 '20 at 07:24

1

$f(x)=\sum\limits_{n=1}^{\infty} nr^{n}-\sum\limits_{n=1}^{\infty} ns^{n}$ where $r=x^{2}$ and $s=\frac {x^{2}} 3$. So you only have to now how to find $\sum\limits_{n=1}^{\infty} nt^{n} $ when $|t|<1$.

For this let $g(t)=\sum\limits_{n=1}^{\infty} t^{n}$. Then $g(t)=\frac t {1-t}$. Also $tg'(t)=\sum\limits_{n=1}^{\infty} nt^{n}$. Can you finish?

answered Aug 07 '20 at 07:24

Kavi Rama Murthy

311,013

Not quite. Could you give a hint on how to take it from there? – Robbert Aug 07 '20 at 08:17
@Robbert Inequalities between metrics are not equivalent to comparability of topologies. This equivalence works only for norms, not fro metrics in general. – Kavi Rama Murthy Aug 07 '20 at 08:24

show that the sum function f is given by F

1 Answers1