Determining the closed form solution for infinite summation

Question

I've been stuck trying to calculate the closed form version of this summation. I've looked around but I still unable to find the answer. I really don't know what to do with the power of x which is the 2i-1.

$$\sum_{i=1}^∞ ix^{2i-1}$$

Answer 1

A different way of getting to the same place. Let $z=x^2$: $$ S = \sum_{i=1}^\infty ix^{2i-1} = x \sum_{i=1}^\infty i\left(x^2\right)^{i-1} = x \sum_{i=1}^\infty iz^{i-1} = x \sum_{i=1}^\infty \frac{d}{dz} \left[z^i\right] = x \frac{d}{dz} \left[ \sum_{i=1}^\infty z^i\right] $$

Answer 2

Let $$S = \sum_{i = 1}^{\infty} ix^{2i - 1} \; \text{for} \; |x| < 1$$

We have \begin{align} S &= x + 2x^3+3x^5 + 4x^7+5x^9 + ...\\ x^2S &=\quad \quad x^3 + 2x^5 + 3x^7 + 4x^9 + 5x^{11} + ...\\ \therefore (1 - x^2)S &= x + \; \, x^3 + \; \,x^5 + \; \, x^7 + \; \, x^9 + ...\\ & = \frac{x}{1 - x^2} \end{align}

Hence,

$$\boxed{S = \frac{x}{(1 - x^2)^2}}$$

Answer 3

Hint

$$\sum_{i=1}^\infty ix^{2i-1}=\frac 12\sum_{i=1}^\infty (2i)\,x^{2i-1}=\frac 12\Bigg[\sum_{i=1}^\infty x^{2i}\Bigg]'$$