# Radius and Interval of Convergence $\sum_{n=0}^{\infty}\frac{7^n}{n!}x^n$

$\displaystyle \sum_{n=0}^{\infty}\frac{7^n}{n!}x^n$

I'm still trying to get the hang of these and feel like I've done something wrong here. After applying the ratio test I end up with:

$\left|7x\right|\lim \limits_{n \to \infty}\left|\frac{1}{n+1}\right|$

That limit is $0$, so does this mean my radius of convergence is $\infty$ and my interval of convergence is $(-\infty, \infty)$?

• Yes, that's right. Note that the series is just $e^{7x}$. – carmichael561 Jul 16 '16 at 1:03
• indeed it is, you might also recognize that series as the McLaurin expansion of $e^{7x}$ – Doug M Jul 16 '16 at 1:03