How to solve this expression: $\sum_{n=1}^\infty\frac1{(x-3)^{2n-1}}$

Question

$$\sum_{n=1}^\infty\frac1{(x-3)^{2n-1}}$$

How to change the given expression to a rational function

Answer 1

Rewrite the sum as $$\frac{1}{x-3}\sum_{n=0}^\infty\left(\frac{1}{(x-3)^2}\right)^n.$$ Now, for any real number $r$ with $-1<r<1$, it is well known that $$\sum_{n=0}^\infty r^n=\frac{1}{1-r}$$ so only if $x<2$ or $x>4$ we can conclude that the desired sum is equal to $$\frac{1}{x-3}\cdot\frac{1}{1-\left(\frac{1}{x-3}\right)^2}=\frac{1}{x-3-\frac{1}{x-3}}=\frac{x-3}{(x-3)^2-1}=\frac{x-3}{x^2-6x+8}.$$ This means that if $2\leq x\leq 4$ then you cannot rewrite the sum as a fraction.

Answer 2

First, make the series zero-based (as the geometric series is zero-based).

$$\sum_{n\ge 1}\dfrac{1}{(x-3)^{2n-1}} = \sum_{n\ge 0} \dfrac{1}{(x-3)^{2n+1}}$$

Next, pull out $\frac{1}{x-3}$:

$$\dfrac{1}{x-3}\sum_{n\ge 0} \left(\dfrac{1}{(x-3)^2}\right)^n = \dfrac{1}{x-3}\cdot \dfrac{1}{1-\tfrac{1}{(x-3)^2}} = \dfrac{x-3}{(x-3)^2-1}, \left|\dfrac{1}{(x-3)^2}\right| < 1$$

Answer 3

This is a geometric series of common ratio $(x-3)^{-2}$ and initial term $(x-3)^{-1}$.

Hence for

$$(x-3)^{-2}<1,$$ the sum converges to

$$\frac{(x-3)^{-1}}{1-(x-3)^{-2}}.$$