Finding Maclaurin series of a function The prompt is to find the Maclaurin series of the function 
$$f(x) = xe^{x^2-1}$$ and evaluate the 100th derivative. At what value of x does this series converge?
We know that $$e^x = \sum \frac{x^n}{n!}$$ 
$$f(x) = \frac{xe^{x^2}}{e}$$
$$\frac{1}{e} f(x) = xe^{x^2} = \sum{\frac{x^{2n+1}}{n!}}$$
For the 100th derivative, we just replace n with 100, giving us 
$$f^{(100)}(0) = \frac{0^{101}}{100!}.\frac{1}{e} = 0$$
I'm not sure how to find the value of x for which the series converges. 
 A: Note that $f$ is an odd function (i.e. $f(-x)=-f(x)$ so the derivatives of even order at $0$ are all zero. Moreover
$$xe^{x^2-1}=\frac{1}{e}\sum_{n=0}^{\infty}\frac{x^{2n+1}}{n!}.$$
Hence the radius of convergence of this power series is the reciprocal of the limit
$$\lim_{n\to +\infty} \left(\frac{1}{en!}\right)^{1/(2n+1)}.$$
Can you take it from here?
A: I think your series is a bit off.  In particular, if you have $$f(x) = \frac{xe^{x^2}}{e}$$ then $$ef(x) = xe^{x^2},$$ not $$\frac{f(x)}{e} = xe^{x^{2}}.$$  You have then $$ef(x) = xe^{x^{2}}=x\sum_{n=0}^{\infty}\frac{x^{2n}}{n!}=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{n!}$$ so $$f(x) = \frac{1}{e}\sum_{n=0}^{\infty}\frac{x^{2n+1}}{n!}.$$
A: $f(x) = e\sum_\limits{n=0}^\infty \frac {x^{2n+1}}{n!}\\
f'(x) = e\sum_\limits{n=0}^\infty \frac {(2n+1)x^{2n}}{n!}\\
f''(x) = e\sum_\limits{n=1}^\infty \frac {(2n+1)(2n)x^{2n-1}}{n!} = e\sum_\limits{n=1}^\infty \frac {(2n+1)!}{(2n-1)!}\frac {x^{2n-1}}{n!}$
Note that n didn't increase in the summation, the first time we took the derivative, but it did, the second.  The is because the lowest powered term was x in the original function, was a constant after taking the derivative.
$f^{(100)}(x) = e\sum_\limits{n=50}^\infty \frac {(2n+1)!}{(2n-100)!}\frac {x^{2n-99}}{n!}$
radius of convergence 
Ratio test:
find x such that 
$\lim_\limits{n\to\infty}\left|\dfrac {\frac {(2n+3)!}{(2n-98)!}\frac {x^{2n-97}}{(n+1)!}}{\frac {(2n+1)!}{(2n-100)!}\frac {x^{2n-97}}{(n)!}}\right| < 1$
$\lim_\limits{n\to\infty}\left|\frac {(2n+3)(2n+2)x}{(2n-99)(2n-98)(n+1)}\right|<1$
As $n$ gets to be very large, the limit approaches $\frac {x}{n}$ and will be less than $1$ for all $x$
The series converges for all x.
