How to find convergence radius and convergence range of $\sum\limits_{n=1}^\infty\frac{(-1)^n}{(2n)!}x^{2n}$?

Question

I can not find the general term of $\sum\limits_{n=1}^\infty\dfrac{(-1)^n}{(2n)!}x^{2n}$ as $a_{2n}$.

I think $a_{2n}$ can be $\dfrac{(-1)^n}{(2n)!}$. If so, how can i find the next term of the series?

Is the next term equal to $a_{2n+1}$ or $a_{2(n+1)}=a_{2n+2}$?

Which one is true?

Answer 1

No mate, here$$a_n =\frac{(-1)^n}{(2n)!}$$ Actually the given series is the expansion of $\cos x -1$