# How do you go from $\sum_{n \geq 3} \frac{1}{3} \big(\frac{1}{2}\big)^{n-2}s^n$ to $\frac{1}{6} s^3 \sum_{n \geq 0}\big(\frac{s}{2}\big)^{n}$?

I'm summing:

$$\sum_{n \geq 3} \frac{1}{3} \bigg(\frac{1}{2}\bigg)^{n-2}s^n$$

Which is just an infinite series, so if I can get it in the form of $$\sum_{n\geq 0}$$, I can use $$s_{\infty}=\frac{a}{1-r}$$.

So I've multiplied it by $$\frac{2^2}{2^2}$$ so that $$1/2$$ and $$s$$ are raised to the same power, i.e. $$\big(\frac{1}{2}\big)^{n-2}s^n=\frac{2^2}{2^2}\big(\frac{1}{2}\big)^{n-2}s^n=2^2\big(\frac{s}{2}\big)^n$$. Then I've summed from $$n=0$$ to $$\infty$$ and subtracted off the extra terms from $$n=0$$ to $$2$$, due to summing from $$n=0$$ instead of $$n=3$$:

$$\frac{2^2}{3} \bigg\{ \sum_{n\geq 0} \bigg(\frac{s}{2}\bigg)^2 -\bigg(\frac{s}{2}\bigg)^0 - \bigg(\frac{s}{2}\bigg)^1-\bigg(\frac{s}{2}\bigg)^2 \bigg\}$$ $$=\frac{4}{3} \bigg\{ \frac{1}{1-\frac{s}{2}} - \frac{s^2+2s+4}{4} \bigg\}$$ $$=\frac{4}{3} \bigg\{ \frac{2}{2-s} - \frac{s^2+2s+4}{4} \bigg\}$$

$$=\frac{4}{3}\bigg\{ \frac{(2)(4)-(2-s)(s^2+2s+4)}{4(2-s)} \bigg\}$$

$$=\frac{1}{3}\bigg\{ \frac{8-(8-s^3)}{(2-s)} \bigg\}$$ $$=\frac{s^3}{3(2-s)}$$

That was all kind of messy. In one fell swoop, my professor just simply wrote down:

$$\sum_{n \geq 3} \frac{1}{3} \bigg(\frac{1}{2}\bigg)^{n-2}s^n = \frac{1}{6} s^3 \sum_{n \geq 0}\bigg(\frac{s}{2}\bigg)^{n}=\frac{s^3}{3(2-s)}$$

This is certainly a cleaner way than my way, however, I'm not even sure what the thought process is here. How exactly can this be reasoned through? How exactly do you go from $$\sum_{n \geq 3} \frac{1}{3} \big(\frac{1}{2}\big)^{n-2}s^n$$ to $$\frac{1}{6} s^3 \sum_{n \geq 0}\big(\frac{s}{2}\big)^{n}$$?

Suggested format for an answer is how I've explained the reasoning behind my messy approach. But feel free to explain in any way you see fit.

Hint : $$\bigg(\frac{1}{2}\bigg)^{n-2}s^n = \bigg(\frac{1}{2}\bigg)^{1}\bigg(\frac{1}{2}\bigg)^{n-3} s^3 s^{n-3}$$ then apply the change of variables in the summand.