So far I have said it diverges due to the comparison test with the harmonic series.

I said $$\frac{3^n}{n^n} < \frac{3^n}{n}< 3^n \frac{1}{n}$$ and as $\frac{1}{n}$ diverges so does the series.

But it also seems the term of the series converges to 0 so that should mean the series is convergent.

closed as off-topic by José Carlos Santos, RRL, Jyrki Lahtonen, user302797, Xander Henderson Dec 8 at 12:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, RRL, Jyrki Lahtonen, user302797, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    You've shown that your series element is smaller than something that diverges, which proves nothing. Either prove that it is bigger than something that diverges or that it is smaller than something that converges (hint: try the latter first). – Ingix Dec 6 at 14:58
  • @Pumpkinpeach What about the given hints? Did you succeed with the resolution? – gimusi Dec 7 at 7:59
  • Yes, the root test was the easiest way I found. Thanks. – Pumpkinpeach Dec 7 at 21:37
  • @Pumpkinpeach It would be useful if you can add your derivation editing your question. Bye – gimusi Dec 8 at 9:16
  • Maybe looking at some similar problems on this site might help you. For example: Is my proof of convergence for $\sum\limits^{\infty}_{n=1} \frac {2^n} {n^n}$ true? I found that one using Approach0. See also: How to search on this site? – Martin Sleziak Dec 8 at 9:28

Hint:

$$\frac{3^n}{n^n} \lt \frac{3^n}{6^n} = \frac{1}{2^n} \text{ when } n \gt 6$$

  • 1
    Do you mean $n^n$ in the denominator? – marty cohen Dec 6 at 15:07
  • @martycohen Yes, thank you. I should have said "and you can use this to show $6.6597 \lt \sum \limits_{n=1}^{\infty} \frac{3^n}{n^n} \lt 6.6755$. It is in fact about $6.6629$" – Henry Dec 6 at 15:26

HINT

What about the root test

$$\sqrt[n]{\left(\frac{3}{n}\right)^n}$$

Edit: note that the fact that $a_n=\frac{3^n}{n^n}\to 0 $ is a necessary but not sufficient condition for convergence.

For any $a > 0$, $\sum a^n/n^n$ converges since $a^n/n^n < 1/2^n$ for $n > 2a$.

$$\begin{aligned}\frac{3^x}{x^x}&\leq\frac{3^x}{x!}\\ \sum_{x=1}^\infty\frac{3^x}{x^x}&\leq\sum_{x=1}^\infty\frac{3^{x}}{x!}\\ &\leq e^{3}-1\\ &\leq20\\ \end{aligned}$$


We could also make this bound tighter with Stirling's approximation. As $\ln(x!)\sim x\ln(x)-x+\frac12\ln(2\pi x)+\ldots$, we may rearrange and exponentiate both sides for

$$\begin{aligned}x^x&\sim\frac{x!}{\exp\left(-x+\frac12\ln(2\pi x)+\ldots\right)}\\ \\ \sum_{k=x_0}^\infty\frac{3^k}{k^k}&\sim\sum_{k=x_0}^\infty\frac{3^k}{k!}\exp\left(-k+\frac12\ln(2\pi k)+\ldots\right)\quad\text{as $x_0\to\infty$}\end{aligned} $$

  • 1
    Nice variation on the usual. – marty cohen Dec 6 at 19:51
  • Thank you, @Marty – Jam Dec 6 at 19:57

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