Infinite Sum of Powers of x over 1-x^2

[Saitama]

Problem:
If $0<x<1$ and
$$A_n=\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\cdots +\frac{x^{2^n}}{1-x^{2^{n+1}}}$$
then find $\displaystyle \lim_{n\rightarrow \infty}A_n$.

Attempt:
I tried to see if it can be converted to a telescoping series but I had no luck. Then, I tried this:
$$\lim_{n\rightarrow \infty} A_n=\sum_{r=0}^{\infty} \frac{x^{2^r}}{1-x^{2^{r+1}}}=\sum_{r=0}^{\infty} \sum_{m=0}^{\infty} x^{2^r(2m+1)}$$
If I can figure out $\displaystyle \sum_{r=0}^{\infty} t^{2^r}$, the sum can be solved easily but I don't see a way to evaluate this.

Any help is appreciated. Thanks!

 

[kaliprasad]

you can put

$\frac{x^{2^k}}{1- x^{2^{k+1}}}$
=$\frac{1}{2}(\frac{1}{1-x^{2^k}}-\frac{1}{1+x^{2^k}}) $

now you get sum of a series - sum of another series and you should be able to proceed

 

[Saitama]

you can put

$\frac{x^{2^k}}{1- x^{2^{k+1}}}$
=$\frac{1}{2}(\frac{1}{1-x^{2^k}}-\frac{1}{1+x^{2^k}}) $

now you get sum of a series - sum of another series and you should be able to proceed

That's what I tried when I thought of making it a telescoping series but I couldn't proceed after that. Writing down a few terms:

$$\frac{1}{2}\left(\frac{1}{1-x^2}-\frac{1}{1+x^2}+\frac{1}{1-x^4}-\frac{1}{1+x^4}+\frac{1}{1-x^8}-\frac{1}{1+x^8}\cdots \right)$$

I don't see where to proceed from here.

 

[kaliprasad]

sorry for last entry which did not help

if you proceed as below you get telescopic sum

$\frac{x^{2^k}}{1- x^{2^{k+1}}}$
= $\frac{x^{2^k}+1-1}{1- x^{2^{k+1}}}$
= $\frac{1}{1- x^{2^k}}-\frac{1}{1- x^{2^{k+1}}}$

 

[Opalg]

Problem:
If $0<x<1$ and
$$A_n=\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\cdots +\frac{x^{2^n}}{1-x^{2^{n+1}}}$$
then find $\displaystyle \lim_{n\rightarrow \infty}A_n$.

[Not sure if this is the same as kaliprasad's method.]

The basic hint is to look at the sum of the first few terms, then see if you can spot a pattern in these sums, guess a formula for the sum of the first $n$ terms, and finally let $n\to\infty$.

[sp]In fact, the first term is $\dfrac x{1-x^2} = \dfrac{(1+x) - 1}{(1+x)(1-x)} = \dfrac1{1-x} - \dfrac1{1-x^2}.$

Now add the second term, to get $\Bigl(\dfrac1{1-x} - \dfrac1{1-x^2}\Bigr) + \dfrac{x^2}{1-x^4} = \dfrac1{1-x} - \dfrac{(1+x^2) - x^2}{(1+x^2)(1-x^2)} = \dfrac1{1-x} - \dfrac1{1-x^4}.$

You should be able to see the pattern emerging ... .[/sp]

 

[kaliprasad]

[Not sure if this is the same as kaliprasad's method.]

The basic hint is to look at the sum of the first few terms, then see if you can spot a pattern in these sums, guess a formula for the sum of the first $n$ terms, and finally let $n\to\infty$.

[sp]In fact, the first term is $\dfrac x{1-x^2} = \dfrac{(1+x) - 1}{(1+x)(1-x)} = \dfrac1{1-x} - \dfrac1{1-x^2}.$

Now add the second term, to get $\Bigl(\dfrac1{1-x} - \dfrac1{1-x^2}\Bigr) + \dfrac{x^2}{1-x^4} = \dfrac1{1-x} - \dfrac{(1+x^2) - x^2}{(1+x^2)(1-x^2)} = \dfrac1{1-x} - \dfrac1{1-x^4}.$

You should be able to see the pattern emerging ... .[/sp]

it is different. mine is based on telescopic sum. your is based on induction ( see the pattern and predict the value)

 

[chisigma]

In my opinion the result obtained by Kaliprasad and Opalg are fully equivalent so that the result seems to be...

$\displaystyle \lim_{n \rightarrow \infty} A_{n} = \lim_{n \rightarrow \infty} \frac{1}{1-x} - \frac{1}{1 - x^{2^{n}}} = \frac{x}{1-x}\ (1)$

Kind regards$\chi$ $\sigma$

 

[Saitama]

sorry for last entry which did not help

if you proceed as below you get telescopic sum

$\frac{x^{2^k}}{1- x^{2^{k+1}}}$
= $\frac{x^{2^k}+1-1}{1- x^{2^{k+1}}}$
= $\frac{1}{1- x^{2^k}}-\frac{1}{1- x^{2^{k+1}}}$

[Not sure if this is the same as kaliprasad's method.]

The basic hint is to look at the sum of the first few terms, then see if you can spot a pattern in these sums, guess a formula for the sum of the first $n$ terms, and finally let $n\to\infty$.

[sp]In fact, the first term is $\dfrac x{1-x^2} = \dfrac{(1+x) - 1}{(1+x)(1-x)} = \dfrac1{1-x} - \dfrac1{1-x^2}.$

Now add the second term, to get $\Bigl(\dfrac1{1-x} - \dfrac1{1-x^2}\Bigr) + \dfrac{x^2}{1-x^4} = \dfrac1{1-x} - \dfrac{(1+x^2) - x^2}{(1+x^2)(1-x^2)} = \dfrac1{1-x} - \dfrac1{1-x^4}.$

You should be able to see the pattern emerging ... .[/sp]

Thanks a lot Opalg and kaliprasad! :)