Power Series, Taylor/Maclaurin and n-th derivative 
a) Find the convergence radius of the power series $$\sum_\limits{n=1}^∞\frac{(n+1)^{n^2}}{3^nn^{n^2}}x^n $$
b)Find all the $x\in \mathbb{R}$ for which the power series $\sum_\limits{n=1}^∞\frac{(x+1)^n}{\sqrt{4^nn}}$ converge.
c)Using the Maclaurin expansion of $e^x$, find the power series of $g(x)=e^{5x^2}$ with center $x_0=0$. Also for every $n\in \mathbb{Z}$ find a formula for $g^{(n)}(0)$.

My work so far:
a) I found $R=\frac{3}{e}$ with $R={(\limsup_{n\rightarrow∞}\sqrt[n]{|a_n|})}^{-1}$.
b) From the ratio test and $n\rightarrow ∞$ we get $|(x+1)\frac{1}{2}|$. So the power series converge for $|(x+1)\frac{1}{2}|<1$. Thus $-3<x<1$ is the interval of convergence of the power series.
c) The power series of $g(x)$ using the Maclaurin expansion of $e^x$ is  ${\sum_\limits{n=0}^ ∞}\frac{5^n}{n!}x^{2n}$
Is my work correct so far?
I don't know what to do for the n-th derivative of $g(x)$. Any help would be appreciated.
 A: The Maclaurin series for $f$ is:
$$f(x)=\sum\limits_{n \in \mathbb{N}} \left.\frac{\mathrm{d}^nf(x)}{\mathrm{d}x^n}\right|_{x=0}\frac{x^n}{n!}$$
I think this will be enough for you to finish the excercise c.  
So, the first few terms of the Maclaurin series are:
$$g^{(0)}(0)\frac{x^0}{0!}+g^{(1)}(0)\frac{x^1}{1!}+g^{(2)}(0)\frac{x^2}{2!}+\dots$$
$$g^{(0)}(0)+g^{(1)}(0)x+g^{(2)}(0)\frac{x^2}{2}+\dots$$
While your function is:
$$5^0\frac{x^{2*0}}{0!}+5^1\frac{x^{2*1}}{1!}+5^2\frac{x^{2*2}}{2!}+\dots$$
$$1+5x^2+\frac{25}{2}x^4+\dots$$
A: Finding the n-th derivative of $g$ is easy, since you have already found the Maclaurin expansion:


*

*Since $g$ is an even function, $g^{(n)}(0)=0$, if $n$ is odd.

*From the Maclaurin expansion, if $n$ is even:


*

*$
\begin{align*}
 g(x)
=\sum_{n=0}^{\infty}\frac{g^{(n)}(0)}{n!}x^n
=\sum_{v=0}^\infty \frac{5^v}{v!}x^{2v}\Rightarrow
x^n=x^{2v}\Rightarrow
n=2v
\end{align*}
$

*$
\begin{align*}
g(x)
=\sum_{n=0}^{\infty}\frac{g^{(n)}(0)}{n!}x^n
=\sum_{v=0}^\infty\frac{5^v}{v!}x^{2v}\Rightarrow
\frac{g^{(n)}(0)}{n!}=\frac{5^v}{v!}\Rightarrow
g^{(n)}(0)
=\frac{5^vn!}{v!}
=\frac{\sqrt{5^n}n!}{\left(\frac{n}{2}\right)!}
\end{align*}
$



A piecewise definition of the n-th derivative of $g$ would be:
$$
g^{(n)}(0)=\begin{cases}\frac{\sqrt{5^n}n!}{\left(\frac{n}{2}\right)!}&n \to even\\ 0&n \to odd\end{cases}
$$
