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

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Answer 1 · 2018-07-30T23:07:42

The interval of convergence is #(-oo, oo)#

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)#.