Using the Cauchy root test

Question

I've got the next series: $$\sum_{n=1}^{\infty} \frac{3^n n!}{n^n} $$ I must to solve it to see if converges or not. So I've got the next process:

Sol (attempt): Let $a_n=\frac{3^n n!}{n^n}$. By the Cauchy root test we've got that: $$\lim_{n\to\infty}\sqrt[n]{a_n} =\lim_{n\to\infty}\sqrt[n]\frac{3^n n!}{n^n} =\lim_{n\to\infty}\sqrt[n] {\left(\frac{3}{n}\right)^n n!} = \lim_{n\to\infty}\frac{3}{n}\sqrt[n]{n!} = \lim_{n\to\infty}\frac{3}{n}(n!)^\frac{1}{n} = 0$$ If $n$ tends to infinity we've got that $\frac{3}{n}=0$ and $ (n!)^\frac{1}{n}=(n!)^0=1$
Thus, the series converges absolutely.

My question is about if I'm using the test in a good way, and if the limit is correct. If there's any error on my process I'd be grateful to know.

Answer 1

We have $$\frac {a_{n+1}}{a_n}= \frac {3(n+1)n^n} {(n+1)^{n+1}} =\frac {3}{ (1+\frac {1}{n})^n }\to \frac {3}{e}>1 \text { as } n\to \infty.$$

Or we can apply the root test using Stirling's Formula: For $n>0$ we have $1-\frac {1}{6n}<n!^{-1}(\frac {n}{e})^n\sqrt {2\pi n}\;<1+\frac {1}{6n}$.

Although for this Q it is sufficient to use a consequence of it that's much easier to prove: $\lim_{n\to \infty}n!\cdot \left(\frac {n}{e}\right)^{-n}=1, $ which implies that $$\lim_{n\to \infty}(a_n)^{1/n}=\frac {3}{e}>1.$$

Answer 2

The ratio test works readily: $$\frac{a_{n+1}}{a_n}=\frac{3^{n+1}(n+1)\not!}{(n+1)^{n+1}}\,\frac{n^{n}}{3^{n}\not{n\not!}}=\frac{3}{\Bigl(\cfrac{n+1}n\Bigr)^n}=\frac{3}{\Bigl(1+\cfrac{1}n\Bigr)^n}\to\frac 3{\mathrm e}>1.$$

Answer 3

I thought that it might be instructive to present a way forward that does not rely on Stirling's Formula. In the following, we will use the inequality

$$\begin{align} \frac1n \sum_{k=1}^n\log(k/n)&>\int_0^1 \log(x)\,dx\\\\ &=-1\tag1 \end{align}$$

since $\log(x)\le 0$ and concave for $x\in (0,1]$. We now proceed.

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Note that we have

$$\begin{align} (n!)^{1/n}&=e^{\frac1n \log(n!)}\\\\ &=e^{\frac1n \sum_{k=1}^n\log(k)}\\\\ &=ne^{\frac1n \sum_{k=1}^n\log(k/n)}\tag2\\\\ &>ne^{-1}\tag3 \end{align}$$

where we used $(1)$ in going from $(2)$ to $(3)$.

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Finally, applying $(3)$ yields

$$\begin{align} \sqrt[n]{\frac{3^n\,n!}{n^n}}&>\frac3e\\\\ &>1\tag4 \end{align}$$

Letting $n\to \infty$ in $(4)$, we find that

$$\lim_{n\to \infty}\sqrt[n]{\frac{3^n\,n!}{n^n}}>1$$

which implies from the root test that the series $\sum_{n=1}^\infty \frac{3^n\,n!}{n^n}$ diverges.

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NOTE: Given $(4)$, we see that the general terms of the series do not approach $0$. Therefore, the root test is unnecessary.

Answer 4

Up to this point it is OK: $$\lim_{n\to\infty}\frac{3}{n}(n!)^\frac{1}{n}$$ However, $(n!)^\frac{1}{n}=(n!)^0=1$ is not true. It is $\infty^0$ type of indeterminate form, so you must be careful. You can use the estimate: $e^n\ge \frac{n^n}{n!} \Rightarrow n!\ge \left(\frac{n}{e}\right)^n$: $$\frac{3}{n}(n!)^\frac{1}{n}\ge \frac3n\cdot \frac{n}{e}=\frac3e>1.$$

Answer 5

Using what you already did, consider $$A=\frac{3}{n}(n!)^\frac{1}{n}\implies \log(A)=\log(3)-\log(n)+\frac{1}{n}\log(n!)$$ and use Stirling approximation $$\log(n!)=n (\log (n)-1)+\frac{1}{2} \left(\log (2 \pi )+\log \left({n}\right)\right)+\frac{1}{12 n}+O\left(\frac{1}{n^3}\right)$$ Replace and simplify to get $$\log(A)=-1+\log(3)+\frac{\log(2\pi n)}{2n}+O\left(\frac{1}{n^2}\right)$$

Answer 6

Here is an answer which uses Stirling's approximation, which differs slightly by the other answers. Originally intended to answer a duplicate of this question, I noticed that none of the answers exactly matched mine, and as the question is still unanswered it may be of sight help... I used the formula the straightforward way without any log simplifications as per the other answer....

By Stirling's approximation, we have

$n!\sim \frac{\sqrt{2\pi n}n^n}{e^n}$

Hence, at the limit to infinity, we can write

$\lim_{n \to \infty} \frac{3^nn!}{n^n}=\lim_{n \to \infty} \frac{3^n}{n^n}\frac{\sqrt{2\pi n}~n^n}{e^n}=\lim_{n \to \infty}\left(\frac{3}{e}\right)^n\sqrt{2\pi n}$

As 3>e, the limit at infinity clearly diverges, hence the entire sum diverges......