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

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 $(-\infty, \infty)$?

  • 2
    $\begingroup$ Yes, that's right. Note that the series is just $e^{7x}$. $\endgroup$ Jul 16, 2016 at 1:03
  • 1
    $\begingroup$ indeed it is, you might also recognize that series as the McLaurin expansion of $e^{7x}$ $\endgroup$
    – Doug M
    Jul 16, 2016 at 1:03

1 Answer 1


Another way to check:the Cauchy-Hadamard theorem simply gives a formula for the radius of convergence: https://en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard_theorem


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.