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I am asked to determine if a series converges or not:
$$\displaystyle\sum\limits_{n=1}^{\infty} \frac{(2^n)n!}{(n^n)}$$
So I'm using the $n$th root test and came up with $\lim_{n \to {\infty}}\frac{2}{n}\times(\sqrt[n]{n!})$ I know that the limit of $\frac{2}{n}$ goes to $0$ when $n$ goes to infinity but what about the $(\sqrt[n]{n!})$?
I thought it might be instructive to present an approach that relies on elementary tools only. To that end, we now proceed.
Note that we can write
$$\begin{align} \log\left(\frac{\sqrt[n]{n!}}{n}\right)&=\frac1n\log(n!)-\log(n)\\\\ &=\frac1n\sum_{k=1}^n\log(k)-\log(n)\\\\ &=\underbrace{\frac1n\sum_{k=1}^n\log(k/n)}_{\text{Riemann Sum for}\,\,\int_0^1 \log(x)\,dx=-1}\\\\ \end{align}$$
Hence, we have
$$\lim_{n\to \infty}\left(\frac{2\sqrt[n]{n!}}{n}\right)=2e^{-1}$$
And we are done!
Tools Used. Straightforward arithmetic and Riemann sums.
>? That is semantic markup for citations, in Markdown. You are supposed to use them only when you are quoting text from someone else, not to make the sentence stand out.
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Feb 25, 2017 at 14:25
Since OP is taking Calculus III, perhaps the ratio test from calculus II is a more suitable way.
Let $a_n = (2^n)n!/n^n$.
\begin{align} \frac{a_{n+1}}{a_n} &= \frac{(2^{n+1})(n+1)!/(n+1)^{n+1}}{(2^n)n!/n^n} \\ &= 2(n+1) \frac{n^n}{(n+1)^n} \frac1{n+1} \\ &= 2 \frac1{\left(1+\frac{1}{n}\right)^n} \\ \end{align}
$$L = \lim_{n \to +\infty} \frac{a_{n+1}}{a_n} = \lim_{n \to +\infty} 2 \frac1{\left(1+\frac{1}{n}\right)^n} = \frac2e < 1 $$
So the series converges.
Alternative method by Stirling's approximation
I type this for fun and to show the power of this formula for $\sum\limits_{n=1}^{\infty} \frac{2^n n!}{n^n}$
Use the root test on $a_n = (2^n)n!/n^n$.
\begin{align} a_n =& \frac{2^n n!}{n^n} \\ \sim& \frac{2^n\sqrt{2\pi n} e^{-n} n^n}{n^n} \\ =& \sqrt{2\pi} \cdot \frac{2^n}{e^n} \cdot \sqrt{n} \end{align}
The limit $1 \le \sqrt{n}^{1/n} \le n^{1/n} \to 1$ as $n \to +\infty$ allows us to recover the ratio $2/e$ in the previous section.
$$L = \lim\limits_{n\to+\infty} a_n^{1/n} = \lim\limits_{n\to+\infty} \frac2e \sqrt{n}^{1/n} = \frac2e$$
Note that we do not need to actually evaluate the limit, we just need to find an upper bound.
Consider that, for $n>m$, $$ \frac{n!}{m!}\leq n^{n-m} $$ As such, if we let $n=6k-a$, where $0\leq a\leq5$, we can observe that $$ n!\leq (6k)!\leq \prod_{i=1}^6 (ik)^k=(720k^6)^k<\left (\frac{20}{6^4}\right)^k(6k)^{6k} $$ Therefore, $$ \sqrt[n]{n!}<\left(\frac{20}{6^4}\right)^{(n+a)/6n}(n+a)^{1+a/n} $$ and thus $$ \frac{2\sqrt[n]{n!}}{n}<2\left(\frac{20}{6^4}\right)^{1/6}\left(\frac{20}{6^4}\right)^{a/6n}(1+a/n)(n+a)^{a/n} $$ Now, as $a$ cannot be larger than 5, we can easily take the limit of each term as $n\to\infty$, to give $$ \lim_{n\to\infty}\frac{2\sqrt[n]{n!}}{n}<2\left(\frac{20}{6^4}\right)^{1/6}\approx 0.997932 $$ Therefore, as the limit is less than 1, it converges.
Note that the $\lim$ in the final line isn't strictly correct notation, as we have not proven that the limit exists. That said, it captures the intent, that for sufficiently large $n$, the expression will be less than $0.997932$.
