5
$\begingroup$

So I am trying to prove whether the following problem converges or diverges?

$$\sum_{n=1}^\infty \left({n\over n+18}\right)^n$$

So I decided to use the Root test.

$$ L = \lim_{n\to \infty}\sqrt[n]{\left({n\over n+18}\right)^n} = \lim_{n\to \infty} {n\over n+18} = 1$$

But that answer is inconclusive, because according to the Root Test, if L $\lt 1$ than the function converges, and if L $\gt 1$, than the function diverges. But my answer is 1. Can someone please suggest some other methods through which I can determine whether the given problem converges or diverges? Thanks Alot

$\endgroup$
  • $\begingroup$ I always tell my students that the first thing they should consider is:"Does the nth term go to 0?" If not, the series cannot converge. Sanath and Andre are giving you great advice. $\endgroup$ – Chris Leary Apr 8 '14 at 4:12
1
$\begingroup$

Use the limit test:

If $\lim_{n\to\infty}\left(\dfrac{n}{n+18}\right)^n$ is not $0$ or does not exist, $\sum \left(\dfrac{n}{n+18}\right)^n$ diverges. Else the test is inconclusive.

Indeed, $\sum \left(\dfrac{n}{n+18}\right)^n$ diverges.

$\endgroup$
  • 2
    $\begingroup$ Dear @Sanath you mean if it's not zero or ..., right? $\endgroup$ – Ehsan M. Kermani Apr 8 '14 at 4:09
  • $\begingroup$ @EhsanM.Kermani Thanks! $\endgroup$ – user122283 Apr 8 '14 at 4:11
1
$\begingroup$

Note that $\left(\frac{n+18}{n}\right)^n=\left(1+\frac{18}{n}\right)^n$. Since $\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=e^x$, we conclude that $\left(\frac{n+18}{n}\right)^n$ has limit $e^{18}$. So the terms of our series have limit $e^{-18}$, and in particular do not approach $0$. It follows that the series diverges.

$\endgroup$

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.