0

I need to prove that the following series converges

$$\sum_{n = 0}^{\infty} \frac{qz^n}{1 - qz^n}$$

for $|q| < 1$, $|z| < 1$.

I thought maybe to use some comparison tests, but I'm not sure if that will work.

Also, note that I have to prove convergence by showing that the sequence of partial sums converge. The partial sum for this series, though it exists (thanks WolframAlpha), is beyond the scope of this class.

MT_
  • 19,603
  • 9
  • 40
  • 81

2 Answers2

1

Using a comparison test is the right idea. You just need to figure out what to compare to.

Since $|q| < 1$ and $z < 1$, we have $|1-qz^n| \ge 1-|qz^n| = 1-|q||z|^n \ge 1-|q| > 0$.

Therefore, $\left|\dfrac{qz^n}{1-qz^n}\right| = \dfrac{|q||z|^n}{|1-qz^n|} \le \dfrac{|q||z|^n}{1-|q|}$.

Can you see what ot compare the terms to? Remember that $\dfrac{|q|}{1-|q|}$ is a constant.

JimmyK4542
  • 54,331
0

Try the limit comparison test:

$$\begin{cases}a_n:=\frac{qz^n}{1-qz^n}\\{}\\b_n=z^n\end{cases}\;\;\implies \frac{|a_n|}{|b_n|}=\frac{|q|}{|1-qz^n|}\xrightarrow[n\to\infty]{}|q|$$

since

$$0\le\frac{|q|}{|1-qz^n|}\le\frac{|q|}{1-|q||z|^n}\xrightarrow[n\to\infty]{}\frac{|q|}{1-|q|\cdot 0}=|q|$$

so assuming $\;q\neq 0\;$ (otherwise the claim is trivial), the limit comparison test tells us our series converges iff $\;\sum b_n\;$ converges, and this last is a geometric series with ration less than one so it does converge.

Timbuc
  • 34,191