Need help with a problem based on Maclaurin series

Question

i've been given the following problem : Write the Maclaurin series of function $$h(x) = x*sin(x^2) , x\in R$$ , then find the 19th derivative of $$h(0)$$ and then prove that the maclaurin series of h(x) converges for x = 1 . I have found the maclaurin series for the function , however i am having trouble with the rest of the problem . Can you please explain to me the steps to solve this problem ? Thanks !

Answer 1

$$\sin x=\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$

$$\sin x^2=\sum_{n=0}^\infty (-1)^n \frac{x^{4n+2}}{(2n+1)!}$$

$$h(x)=\sum_{n=0}^\infty (-1)^n \frac{x^{4n+3}}{(2n+1)!}$$

To find $h^{(19)}(0)$, you only need to take the term $x^{19}$ from the series expansion of $h(x)$, because all terms $x^{k},~k\le 18$ vanish after taking 19-th derivative. And all terms $x^{k},~k\ge 20$ vanish once plug in $x=0$ after you take the 19-th derivative. So we just let $n=4$, and pick up:

$$\frac{x^{19}}{9!}$$

After do the derivative, we get:

$$h^{(19)}(0)=\frac{19!}{9!}$$

Next, at $x=1$, we have the series:

$$\sum_{n=0}^\infty (-1)^n \frac{1}{(2n+1)!}$$

You can prove it is absolutely convergent by using ratio test.