Convergence area and series representation

Question

Let $$\sum _{k=0}^{\infty }\frac{(-1)^k x^{2k}}{(2k+2)!}$$$x$ is real.

Find the convergence area and series representation.

This one is strange for me. First off I had to find the radius of convergence. Using the ratio test for power series I found it to be $R=\infty$.

I would have thought that would simplify it for me but now I have no idea on how to argue for the area of convergence and series representation. I am thinking the area is a disc at origo with a radius that is strictly less than $+\infty$ and strictly greater than $-\infty$.

José Carlos Santos · Answer 1 · 2021-06-03T20:42:47.683

0

Yes, it follows from the ratio test that the radius of convergence is $\infty$. And what that means is that that series converges everywhere.

Besides, since$$(\forall x\in\Bbb R):\cos(x)=\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}x^{2k},$$you have, for every $x\in\Bbb R$,\begin{align}\frac{1-\cos(x)}{x^2}&=-\sum_{k=1}^\infty\frac{(-1)^k}{(2k)!}x^{2k-2}\\&=-\sum_{k=0}^\infty\frac{(-1)^{k+1}}{(2k+2)!}x^{2k}\\&=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+2)!}x^{2k}.\end{align}

edited Jun 03 '21 at 20:42

answered Jun 03 '21 at 18:04

José Carlos Santos

427,504

What about the series representation? – Jun 03 '21 at 20:22
What is that? The sum of the series? – José Carlos Santos Jun 03 '21 at 20:27
It is the sum function. I.e: https://math.stackexchange.com/questions/2823569/sum-function-for-power-series – Jun 03 '21 at 20:35
I've added it to my answer. – José Carlos Santos Jun 03 '21 at 20:42
I see! makes a lot of sense. For the area of convergence would it be the disc at origo given by $\left { x\in\mathbb{R}:|x|<\infty \right }$? – Jun 03 '21 at 21:38
No. The region of convergence is simply $\Bbb R$. As I wrote in my answer, the series converges everywhere. – José Carlos Santos Jun 03 '21 at 21:42

Convergence area and series representation

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