supremum and infimum: $\frac{n^n}{n!^2}$ So I have this set and I need to find a sup and inf.
$$A=\{\frac{n^n}{n!^2}:n=1,2,3...\}$$
I'd like to know if the part of proof that I have already done is good and I need help with the other part.
I want to check if the series $\frac{n^n}{n!^2}$ is monotonic.
$$\frac{(n+1)^{n+1}}{(n+1)!^2}-\frac{n^n}{n!^2}=\frac{(n+1)^n(n+1)}{(n!(n+1))^2}-\frac{n^n}{n!^2}=$$
$$=\frac{(n+1)((n+1)^n-n^n(n+1))}{n!^2(n+1)^2}=\frac{((n+1)^n-n^n(n+1))}{n!^2(n+1)}$$
$n>0$ so $(n+1)>0$ and $(n+1)^n \ge n^n(n+1)$. So $\frac{((n+1)^n-n^n(n+1))}{n!^2(n+1)}\ge 0$. That means the series is decreasing so it has a supremum. For $n=1$
$$\frac{n^n}{n!^2}=1=\sup A$$ 
$n \in \Bbb N$ so $0$ must be the lower bound. I have to show that $0$ is infimum. So
$$\forall \epsilon \exists n:\frac{n^n}{n!^2}\le0+\epsilon$$
And I think that I have to show this $n$, but I don't know how to do this. I'm stuck.
And sorry for my poor english.
I think that limits may be helpful there. I'd like to know the 2 ways of solving this: with limits and without limits.
 A: To derive the limit you cant take $L = e ^{\log L} = e^{n \log n - 2 \log n!} = e^{n \log n -2 \sum_k \log k}$
Now you can just bound the log-sum with the integral. Keep in mind $\int_{1}^{n} \log x dx < \sum_k \log k < \int_{1}^{n+1} \log x dx$. Can you handle from here? 
A: We have:
$$\frac{a_{n+1}}{a_n} = \left(\frac{(n+1)^{n+1}}{(n+1)!(n+1)!}\right)\left(\frac{n!n!}{n^n}\right) = \frac1{(n+1)}\left(1+\frac1n\right)^n. $$
Since $$\lim_{n\to\infty}\left(1+\frac1n\right)^n=e,$$
it follows that $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=0$. So in fact by the ratio test, the series
$$\sum_{n=1}^\infty \frac{n^n}{n!n!} $$
converges absolutely.
A: You can directly show $n!^2$ eventually gets much bigger than $n^n$. $n^n$ is a product of $n$ terms of size $n$. You can split $n!$ (roughly) into $n/2$ terms that are at least $n/2$ and the rest of the terms have enough factors of $2$ that you can pair them with the other terms to make $n/2$ terms at least $n$.
Then $n!^2$ has $n$ factors at least $n$, so it's already got to $n^n$. You just need to find enough terms left over to ensure that it gets much bigger.
A: Here is an elementary way: it only uses a refined version of Bernoulli's inequality.
Let $u_n=\dfrac{n^n}{(n!)^2}$. We first show $(u_n)$ is a decreasing sequence:
$$\frac{u_{n+1}}{u_n}=\frac{(n+1)^{n+1}}{\bigl((n+1)!\bigr)^2}\cdot\frac{(n!)^2}{n^n}=\Bigl(\frac{n+1}{n}\Bigr)^n\cdot\frac1{n+1}$$
Now it is well known that the first factor tends to $\mathrm e$ as $n$ tends to $\infty$, and is actually bounded by 4. We'll prove this claim in an elementary way in a moment.
So $\,\dfrac{u_{n+1}}{u_n}<\dfrac4{n+1}\le 1$ if $n\ge 3$, and as it is also equal to $1$ if $n=1,2$, we've proved the sequence is nonincreasing for all $n\ge 1$.
For $n>1$ we can write $u_n\le \dfrac{4u_{n-1}}{n}\le \dfrac{4u_1}{n}=\dfrac4n$, which will be ${}<\varepsilon\,$ if $\,n>\dfrac\varepsilon4$.
Proof of the claim
$$\Bigl(\frac{n+1}n\Bigr)^n<\Bigl(\frac{n+1}n\Bigr)^{n+1}.$$
The latter is a decreasing sequence: denote it $a_n$. Indeed
$$\frac{a_n}{a_{n-1}}=\frac{(n+1)^{n+1}}{n^{n+1}}\cdot\frac{(n-1)^n}{n^n}=\Bigl(\frac{n^2-1}{n^2}\Bigr)^n\,\frac{n+1}{n}= \Bigl(1-\frac{1}{n^2}\Bigr)^n\,\Bigl(1+\frac1n\Bigr)$$
Since $\dfrac1{n^2}>-1$, we can apply Bernoulli's inequality:
$$\Bigl(1-\frac{1}{n^2}\Bigr)^n\le 1-n\,\frac{1}{n^2}+\frac{n(n-1)}2\frac{1}{n^4}<1-\frac{1}{n}+\frac{1}{2n^2}$$
hence $$\frac{a_n}{a_{n-1}}<\Bigl(1-\frac1n+\frac{1}{2n^2}\Bigr)\Bigl(1+\frac1n\Bigr)=1-\frac1{2n^2}+\frac1{2n^3}\le1\quad\text{for all}\quad n> 1.$$
Thus $\,a_n<a_1=4$ which proves $(a_n)$ is bounded by $4$.
