$(a_k)$ 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

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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.

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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.

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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.

Answer 1

Answer

$|a_k|\le M$ for all $k\in\mathbb N$ and $|x|\lt 1$ $\Leftrightarrow |x^k|\lt 1$

$\Rightarrow |a_k||x^k|\le M|x^k|$

$\Rightarrow \sum_{k=1}^{\infty}|a_{k}x^{k}| \le M\sum_{k=1}^{\infty}|x|^{k}$

Since $|x|\lt 1$ $\Rightarrow \sum_{k=1}^{\infty}|x|^{k}$ converges to some $p$, since it satisfies the converging properties of a geometric series.

$\Rightarrow \sum_{k=1}^{\infty}|a_{k}x^{k}| \le Mp$

$\Rightarrow \sum_{k=1}^{\infty}|a_{k}x^{k}|$ converges

$\Rightarrow \sum_{k=1}^{\infty}a_{k}x^{k}$ converges absolutely

I reckon this is a better approach.

Answer 2

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 $n\rightarrow\infty$.