Let$\{a_n\}_{n=0}^{\infty}$ be bounded. Then $\sum_0^{\infty} z^na_n$ converges for $|z|<1$

Question

How to prove:

Let $\{a_n\}_{n=0}^{\infty}$ be bounded. Prove $\sum_0^{\infty} z^na_n$ converges for $|z|<1$

So far I prove that the partial sums form a Cauchy sequence, i.e. $\{S_k\}_{k=0}^{\infty}$ is Cauchy.

How do I proceed?

---

I think I should suppose the limit is $0$ at first, and then prove that there exists an integer $K$ such that for every $\epsilon$, when $k>K$, $|S_k-0|<\epsilon$.

But where could I use the fact that I have proved (i.e. partial sums form a Cauchy)?

Any hint?

Answer 1

$\left | a_0 za_1 z^2a_2 ... \right |\quad< \left | a_0 \right | + \left | z \right | \left | a_1 \right | + \left | z^2 \right |\left | a_2 \right | + \dots$
since the sequence is bounded, there exists an upper
bound $(a)$ $\Rightarrow$ $LHS < \left | a \right |(1 + \left | z \right | + \left | z^2 \right |.....)\\ LHS < \left | a \right |(1/1-\left | z \right |)$
which is a finite quantity. Hence the series converges.

you don't have to use Cauchy sequence