# Writing a power series in closed form

Could someone please explain how to write this in closed form, as I have an additional variable in my series?

My goal is to write $$\sum_{n=1}^{\infty} (4 - \frac{x}{3})^{n}$$ in a closed form.

I think this is different than a lot of the other questions out there, so I thought I'd make it one on its own.

• it is a geometric series
– Nick
Oct 6 '19 at 15:30
• What have you tried? What happens if you just treat it like a standard geometric series? Don't be intimidated by the fact that the ratio between successive terms is a function of $x$. Oct 6 '19 at 15:31

The first thing I would do is let $$u= 4- \frac{x}{3}$$. Then the series becomes $$\sum_{n=1}^\infty u^n= \sum_{n= 0}^\infty u^n- 1$$. That last sum is the "geometric series" that converges to $$\frac{1}{1- u}$$ so the original sum is $$\frac{1}{1- u}- 1= \frac{u}{1- u}= \frac{4- \frac{x}{3}}{1- 4+ \frac{x}{3}}= \frac{4- \frac{x}{3}}{\frac{x}{3}- 3}$$. Multiply both numerator and denominator by 3 to get $$\frac{12- x}{x- 9}$$.
Hint: Set $$u=4-\dfrac x3$$ and use the formula $$\sum_{n\ge 1}u^n=\frac u{1-u}\quad\text{ for all u such that } |u|<1.$$