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.
Best Answer
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?