I am a bit confused as to whether I am doing this question correctly.
Firstly, we have defined the radius of convergence of a power series centered at a $$\sum_{n=0}^{\infty} a_n(x-a)^n$$
to be the positive real number $R$ such that the power series converges uniformly on the interval $(a-R,a+R)$ and for $x \lt a – R$, $x \gt a + R$, the series does not converge.
So, we must show that the radius of convergence of the Taylor series of $e^x$ centered at 0 is infinite. Initially, I answered showing it converged using the ratio test, however, I believe this is wrong as it would only show that the series converges, not necessarily uniformly, correct?
I was wondering if it would be correct to show that for the series $f_n(x) = \frac{x^n}{n!}$ that the series $\sum_{n=0}^{\infty} \frac{x^n}{n!}$ converges absolutely and uniformly on the bounded interval $(-a,a)$ for any $a \in \Bbb{R}$ using the Weierstrasse M-test (by taking $M_n = \frac{a^n}{n!}$ and showing that the sum of this series converges by the ratio test) and then conclude that because I can choose any $a \in \Bbb{R}_0^+$ for which the series converges uniformly, the radius of convergence is infinite (ie. I can choose an interval of any size for the M-test), or do I have to show that the series converges uniformly specifically on the interval $(-\infty,\infty)$?
Thanks for your help!
Best Answer
Hint: $R = \displaystyle \lim_{n \to \infty} \dfrac{|a_n|}{|a_{n+1}|}$, and $a_n = \dfrac{1}{n!}$.