Answer
The radius of convergence: $\infty$
The interval of convergence: $(-\infty,\infty)$
Work Step by Step
Let $a_n=(-1)^n\frac{x^{2n+1}}{(2n+1)!}$.
It can be checked that $\frac{a_{n+1}}{a_n}=\frac{-x^2}{(2n+3)(2n+2)}.$
We know $\lim \frac{-1}{(2n+3)(2n+2)}=0$.
Find the interval of convergence.
$\lim |\frac{a_{n+1}}{a_n}|=\lim |\frac{-x^2}{(2n+3)(2n+2)}|=0\cdot |x^2|=0$
Since $\lim|\frac{a_{n+1}}{a_n}|<1$ for any $x$, the radius of convergence is $\infty$ and the interval of convergence is $(-\infty,\infty)$.
