# Does the series $\frac{3^n}{n^n}$ converge or diverge? [closed]

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 HendersonDec 8 at 12:29

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• 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:

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

• 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}

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