Confirmation of Proof: $\forall a,b \in\mathbb{R}, \ \sum_{n=0}^\infty \frac{a}{b^n} = \frac{ab}{b-1}\Leftrightarrow \frac b2\in(1, a].$

Question

I have created the following conjecture:

$$\forall a,b \in\mathbb{R}, \ \sum_{n=0}^\infty \frac{a}{b^n} = \frac{ab}{b-1}\Leftrightarrow \frac b2\in(1, a].$$

However, I cannot prove this.

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My Attempt:

I applied the ratio test to see whether or not the sum converges.

Let $r_n = \dfrac{a}{b^n}$ then I evaluated the following limit:

$$L = \lim_{n\to\infty}\frac{r_{n+1}}{r_n} = \lim_{n\to\infty}\left(r_{n+1}\cdot\frac{1}{r_n}\right) = \lim_{n\to\infty}\left(\frac{a}{b^n}\cdot\frac{b^{n+1}}{a}\right) = \lim_{n\to\infty}\frac{b^{n+1}}{b^n} = \lim_{n\to\infty}b = b.$$

I obviously did something wrong; rather, I feel I need to do something like

$$\lim_{n\to\infty}\frac{a}{b^n} = a\cdot\lim_{n\to\infty}\frac{1}{b^n} = a\cdot 0 = 0,\tag*{$\because b > 1$}$$

but I am not sure. I mean, if $L = 0$, then $L < 1$ and therefore the series is absolutely convergent.

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Could somebody please help me?

Thank you in advance.

Answer 1

$\sum_{n=0}^{\infty}\frac{a}{b^n}$ is a geometric series, hence it converges if and only if $|\frac{1}{b}|<1$

If you evaluate the sum you get $\sum_{n=0}^{\infty}\frac{a}{b^n}=a\frac{1}{1-\frac{1}{b}}=\frac{ab}{b-1}$

Your ratio test gives $\frac{1}{b}$ rather than $b$ - this says that the series converges for $|\frac{1}{b}|<1$