0
$\begingroup$

Determine whether the series

$\sum\limits_{n=1}^\infty \frac{π^n+n}{3^n+n!} $

is convergent or divergent.

Is a positive series and I can see that for $x\to \infty $

$\frac{π^n}{n!}$

and I don't know how to use the comparison test

$\endgroup$
3
$\begingroup$

It converges. $$ 0\le \frac{\pi^n + n}{3^n+n!}\le \frac{2\pi ^n}{n!} $$ and $$ \lim_{n\to\infty}\frac{\frac{2\pi^{n+1}}{(n+1)!}}{\frac{2\pi^n}{n!}}=0, $$ so $\sum_{n=1}^{\infty}\frac{2\pi^n}{n!}$ converges by ratio test. Therefore, given series converges by the comparison test.

$\endgroup$
  • 2
    $\begingroup$ you even can compute $\sum_{n=1}^\infty 2\frac{\pi^n}{n!} = 2 (e^{\pi}-1)$ $\endgroup$ – MJ73550 Apr 22 '16 at 11:04
  • $\begingroup$ @MJ73550 Thanks for giving exact value of that. It is from $e^x =\sum_{n=0}^{\infty} \frac{x^n}{n!}$, right? $\endgroup$ – choco_addicted Apr 22 '16 at 11:06
  • $\begingroup$ @choco_addicted Yep it is. $\endgroup$ – Eff Apr 22 '16 at 11:07
  • $\begingroup$ @choco_addicted Why the numeratior is $2\pi^n$ ? thanks $\endgroup$ – GiovanS Apr 22 '16 at 11:09
  • 1
    $\begingroup$ I replaced $n$ with $\pi^n$, because $n < \pi^n$ is true for $n\in \mathbb{N}$. $\endgroup$ – choco_addicted Apr 22 '16 at 11:11

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.