Determine that $\Sigma^{\infty}_{n=1}\frac{n!}{n^n}$ is convergent.

Question

This is the question from my book.

Is the series $\Sigma^{\infty}_{n=1}\frac{n!}{n^n}$ convergent or divergent?

Actually I know the answer is convergent by plotting but I can't complete the proof.

This is my workings:

$$\lim_{n->\infty}\frac{U_{n+1}}{U_n}=p$$

$$p<1 (convergent) ,p>1(divergent)$$ $$=\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}}$$ $$=\frac{(n+1)n!}{(n+1)^n(n+1)}.\frac{n^n}{n!}$$ $$=\frac{n^n}{(n+1)^n}$$ $$=\left(\frac{n}{n+1}\right)^n$$ This is where I am stuck. Where do I go from here?

Answer 1

Note that

$$\left(\frac n{n+1}\right)^n=\left(\left(1+\frac1n\right)^n\right)^{-1}\to e^{-1}<1$$

Answer 2

You are correct so far. Now note that $\frac{n}{n+1} = 1-\frac{1}{n+1}$, and use the fact that $(1+\frac{x}{n})^n \to e^x$ as $n \to \infty$.

Answer 3

In fact, we can write $$\lim_{n \to \infty} (\frac{n}{1+n})^n = \lim_{n\to \infty} \frac{1}{(1+\frac{1}{n})^n} = \frac{1}{e} < 1$$ Hence, by ratio test the series is convergent. Hope it helps.

Answer 4

You can use Stirling's approximation

to obtain $$ \frac{n!}{n^n} \leq e \frac{n^{n+1/2} e^{-n}}{n^n} = e \frac{\sqrt{n}}{e^n} . $$

Answer 5

The last limit you wrote is known to be equal to $\frac{1}{e}$, so the series converges.

If you want another way to do this, you can use A.M.-G.M. inequality to show that $n!\leq \left(\frac{n+1}{2}\right)^n$, and thus $$\frac{n!}{n^n}\leq \left(\frac{n+1}{2n}\right)^n\leq \left(\frac{3}{4}\right)^n$$for $n\geq 2$; hence the series is convergent by comparison.

Answer 6

You are trying to use the Ratio Test, and your limit converges to $e^{-1}<1$. You can use L'Hospital's Rule to show this. Some traditions even define $e^{-1}$ to be a limit like this.

But here is an alternative. I claim $$\frac{n!}{n^n}<\frac{1}{n^2}$$ for sufficiently large $n$. This is equivalent to $$n!<n^{n-2}$$ $$6\left(\overbrace{4\cdots n}^{n-3\text{ factors}}\right)<n\left(\overbrace{n\cdots n}^{n-3\text{ factors}}\right)$$ Once $n>6$, then what remains to consider on the left is a product of $n-3$ factors. What remains to consider on the right side is also a product of $n-3$ factors, all of which are greater than $1$ and at least as large as any factor on the left.

So you can use comparison with $\sum\frac{1}{n^2}$, a nice $p$-series.

Answer 7

Take $$\log \mathcal{l}=\lim_{n\to\infty}n\log\left(\frac{n}{n+1}\right),$$ where $\mathcal{l}=\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n$, and this remarks make sense if you don't remember how get this limit directly as was showed in a previous answer.

Then write it as a evaluation $\frac{0}{0}$, that is $$\log \mathcal{l}=\lim_{n\to\infty}\frac{\log\left(\frac{n}{n+1}\right)}{1/n},$$ using L'Hôpital's rule, since after you do the calculations, you can get $$\lim_{n\to\infty}\frac{\log\left(\frac{n}{n+1}\right)}{1/n}=\lim_{n\to\infty}\frac{\frac{1}{n(n+1)}}{\frac{-1}{n^2}}=-1.$$

Then by ratio test the series is convergent, since $\mathcal{l}=e^{-1}<1.$