Sum of power series: $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+2)!} x^{n}$

Question

I am asked to find the sum function of the series: $\sum_{n=0}^\infty \frac{(-1)^n}{(2n+2)!} x^{n}$.
I know that $\cos(x)=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{n}$, but I dont know how to divide this series by $(2n+1)(2n+2)$

Help is appreciated!

It is more or less the series for $\cos x$ with some terms missing. — vonbrand, Jun 05 '21 at 01:52
Relabel $n=m-1$ and your series for $\cos x$ is not quite right — Empy2, Jun 05 '21 at 01:53
You should have a $2n$ in the exponent of $x$ for your Maclaurin expansion for $\cos$ — Matthew H., Jun 05 '21 at 01:54
Oh yeah, thank you, I'm sorry for the mistake, I have been thinking about this problem for far too long :) — Saim HQ, Jun 05 '21 at 02:06
Please, do not change the question after receiving an answer. May I ask you to reedit the post with the orginal settings ? Thanks — Claude Leibovici, Jun 05 '21 at 02:25
I've reedited it back, sorry for messing that up, I'm quite new to this forum! — Saim HQ, Jun 05 '21 at 02:36
@SaimHQ. No problem ! If you want to address a comment to a user, start you message with @username. Cheers :-) — Claude Leibovici, Jun 06 '21 at 08:50

score 1 · Accepted Answer · answered Jun 05 '21 at 01:56

1

Hint

To make the two series "almost" similar, let $x=y^2$ $$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+2)!} x^n=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+2)!} y^{2n}=\frac 1 {y^2}\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+2)!} y^{2n+2}$$

answered Jun 05 '21 at 01:56

Claude Leibovici

260,315

The OP has edited the question so that it no longer has $x^n$ but $x^{2n}$ instead. – shoteyes Jun 05 '21 at 02:18
I think the argument still counts with no need for the substitution, but is the last realization that the series is $(cos(x) - 1)/x^2$? – Saim HQ Jun 05 '21 at 02:25

Sum of power series: $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+2)!} x^{n}$

1 Answers1