Sum to infinite terms if $|x|<1$

Question

Sum to infinite terms if $|x|<1$

$$S=\dfrac{1}{\left(1-x\right)\cdot\left(1-x^3\right)}+\dfrac{x^2}{\left(1-x^3\right)\cdot\left(1-x^5\right)}+\dfrac{x^4}{\left(1-x^5\right)\cdot\left(1-x^7\right)}+...............$$

My attempt is as follows:

$$T_n=\dfrac{x^{\left(2n-2\right)}}{\left(1-x^{\left(2n-1\right)}\right)\cdot\left(1-x^{\left(2n+1\right)}\right)}$$

$$\sum_{n=1}^{\infty}T_n=\sum_{n=1}^{\infty}\dfrac{x^{\left(2n-2\right)}}{\left(1-x^{\left(2n-1\right)}\right)\cdot\left(1-x^{\left(2n+1\right)}\right)}$$

$$\sum_{n=1}^{\infty}T_n=\dfrac{1}{x\cdot\left(1-x^2\right)}\cdot\sum_{n=1}^{\infty}\dfrac{x^{\left(2n-1\right)}\cdot\left(1-x^2\right)}{\left(1-x^{\left(2n-1\right)}\right)\cdot\left(1-x^{\left(2n+1\right)}\right)}$$

$$\sum_{n=1}^{\infty}T_n=\dfrac{1}{x\cdot\left(1-x^2\right)}\cdot\sum_{n=1}^{\infty}\dfrac{\left(1-x^{\left(2n+1\right)}\right)-\left(1-x^{\left(2n-1\right)}\right)}{\left(1-x^{\left(2n-1\right)}\right)\cdot\left(1-x^{\left(2n+1\right)}\right)}$$

$$\sum_{n=1}^{\infty}T_n=\dfrac{1}{x\cdot\left(1-x^2\right)}\cdot\left(\sum_{n=1}^{\infty}\dfrac{1}{\left(1-x^{\left(2n-1\right)}\right)}-\dfrac{1}{\left(1-x^{\left(2n+1\right)}\right)}\right)$$

$$\sum_{n=1}^{\infty}T_n=\dfrac{1}{x\cdot\left(1-x^2\right)}\cdot\left(\dfrac{1}{1-x}-\dfrac{1}{1-x^3}+\dfrac{1}{1-x^3}-\dfrac{1}{1-x^5}+............\right)$$

All terms will cancel out to infinity except the first term.

$$\sum_{n=1}^{\infty}T_n=\dfrac{1}{x\cdot\left(1-x^2\right)}\cdot\dfrac{1}{1-x}$$

But surprisingly answer given is

$$\sum_{n=1}^{\infty}T_n=\dfrac{1}{\left(1-x^2\right)}\cdot\dfrac{1}{1-x}$$

I checked multiple times but not able to understand the mistake.

Answer 1

$$\lim_{N\to \infty} \sum_{n=1}^N\left(\frac{1}{1-x^{2n-1}} - \frac{1}{1-x^{2n+1}}\right) = \lim_{N \to \infty} \left(\frac{1}{1-x} - \frac{1}{1-x^{2N+1}}\right) = \frac{1}{1-x} - 1 = \frac{x}{1-x}.$$