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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.

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    $\begingroup$ it is a geometric series $\endgroup$
    – Nick
    Oct 6 '19 at 15:30
  • $\begingroup$ 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 $x$. $\endgroup$
    – mjqxxxx
    Oct 6 '19 at 15:31
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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 $\frac{12- x}{x- 9}$.

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Hint: Set $u=4-\dfrac x3$ and use the formula $$\sum_{n\ge 1}u^n=\frac u{1-u}\quad\text{ for all $u$ such that } |u|<1.$$

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