$(a_k)$ is a sequence with $|a_k|\le M$ for all k . Show that with $|x|\lt 1$ the series $f(x)=\sum_{k=1}^{\infty}a_{k}x^{k}$ converges. 
Question
Let $(a_n)_{n\in\mathbb N}$ be a sequence of real numbers with $|a_k|\le M$ for all $k \in\mathbb N$. Show that for each $x\in\mathbb R$ with $|x|\lt 1$ the series $f(x)=\sum_{k=1}^{\infty}a_{k}x^{k}$ converges.

Proof
We know that:
$|a_k|\le M$ for all $k\in\mathbb N$ and $|x|\lt 1$
$\Rightarrow |a_{k+1}|\le M $ for all $k\in\mathbb N$ and $|x^k|\lt 1$ $\Leftrightarrow |x^{k+1}|\lt 1$
$\Rightarrow |a_{k}x^k|\lt M$ and $|a_{k+1}x^{k+1}|\lt M$
$\Rightarrow |x\frac{a_{k+1}}{a_{k}}|\lt1$
Ratio test:
$\lim_{k\to\infty}\frac{|a_{k+1}x^{k+1}|}{|a_{k}x^k|}$=$\lim_{k\to\infty}|\frac{xa_{k+1}}{a_{k}}|$=$\lim_{k\to\infty}|x\frac{a_{k+1}}{a_{k}}|$$\lt1$
$\Rightarrow \sum_{k=1}^{\infty}|a_{k}x^{k}|$ converges
$\Rightarrow \sum_{k=1}^{\infty}a_{k}x^{k}$ converges absolutley.

Comments
This is a question which was posed to me on my analysis course. Unsure whether I have answered it, having difficulty learning from lecture notes at the moment (my university has closed) :(
Would be great if anyone can refer me to the right theorems or alternatively send their own proofs.

You may want to try the root test instead for then
$\sqrt[n]{|a_n x^n|}\leq \sqrt[n]{M}|x|\xrightarrow{n\rightarrow\infty}|x|$
and so, convergence occurs for all |x|<1.
Check the wikipedia links to root and ration test. Also, you nee to know that for any number $a>0$, $\sqrt[n]{a}\rightarrow1$ as $n\rightarrow\infty$.
