Finding the radius of convergence of $\sum\limits_{k=1}^{\infty}{(a-b)(-b)^{n-1}x^n}$

Question

Here is the summation:

$$\sum\limits_{n=1}^{\infty}{(a-b)(-b)^{n-1}x^n}$$

Now, I tried

$$\lim_{n \to \infty}{\left|\frac{u_{n+1}}{u_n}\right|}$$

which eventually gave me

$$\lim_{n \to \infty}{\left|\frac{-b}{x}\right|}=\left|\frac{-b}{x}\right|$$

So solving

$$\left|\frac{-b}{x}\right|<1$$

should give me the radius of convergence, right? But I tried solving it and I get a strange answer. The answer in the book is

$$\frac{1}{\left|b\right|}$$

What am I doing wrong?

Answer 1

With the quotient rule (Why? Beats me...but whatever):

$$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{(-b)^nx^{n+1}}{(-b)^{n-1}x^n}\right|=|bx|<1\iff |x|<\frac1{|b|}$$

With Cauchy-Hadamard formula:

$$\frac1R=\lim\sup_{n\to\infty}\sqrt[n]{|a_n|}=\lim\sup_{n\to\infty}\sqrt[n]{|(-b)^{n-1}|}=\lim\sup_{n\to\infty}\frac{\sqrt[n]{|-b|^n}}{\sqrt[n]{|-b|}}=|b|$$

The convergence radius is thus $\;R=\cfrac1{|b|}\;$

Answer 2

Your limit is $|b||x|$.

if this limit is $<1$ which is equivalent to

$|x|<\frac{1}{|b|}$,

then the series is convergent.

and if this limit is $>1$ which means

$|x|>\frac{1}{|b|}$, it diverges.

So, the radius of convergence is

$$\frac{1}{|b|}$$