What is the interval of convergence for the Taylor series of f(x)=e^(x^2)?

1 Answer
Jul 30, 2018

The interval of convergence is (-oo, oo)

Explanation:

Recall that the Taylor series of e^t is

e^t = sum_{n=0}^{oo}t^n/(n!)

For this function, substitute color(red)(t = x^2).

e^color(red)(x^2) = sum_{n=0}^{oo}(color(red)(x^2))^n/(n!)

Apply the ratio test.

r = lim_{n->oo}|a_{n+1}/a_n|

If r < 0, then the series is absolutely convergent.

r = lim_{n->oo}|{(x^2)^{n+1}}/{(n+1)!} * {n!}/(x^2)^n|

r = lim_{n->oo}|((x^2)^n*x^2)/{(n+1) * n!} * {n!}/(x^2)^n|

r = lim_{n->oo}|(x^2)/(n + 1)|

And since x^2 is independent of the limit:

r = x^2 * lim_{n->oo}|1/(n + 1)|

r = x^2 * 0

r = 0

Regardless of the value of x, the Taylor series absolutely converges. The interval of convergence then must be (-oo, oo).