Infinite sum to fraction

Question

I have the following infinite sum: $$\sum_{n=0}^{\infty}(n+1)^2 \cdot z^n$$

Could you help me how can I convert it to the fraction form?

$$-\frac{z(z+1)}{(z-1)^3}$$

(when $|z| < 1$)

Answer 1

Recall: $$S(z) = \sum_{n=0}^\infty z^{n} = \frac 1{1-z}\ \ \text{for }|z|< 1$$ Your summation is equivalent to: $$\frac{d}{dz}\left(z\cdot\left(\frac{d}{dz}\left(z\cdot S(z)\right)\right)\right) = \frac d{dz}\left(z\cdot \left(\sum_{n=0}^\infty(n+1)z^n\right)\right) $$$$= \frac d{dz}\left(\sum_{n=0}^\infty(n+1)z^{n+1}\right) = \sum_{n=0}^\infty(n+1)^2z^n$$

If you perform the same layered transformation on $1/(1-z)$ you should arrive at your desired result.

Answer 2

Start with $\sum z^n = \dfrac{1}{1-z}$

1) Multiply both sides by z

2) Differentiate

3) Multiply by z

4) Differntiate