Is this convergent or diverges to infinity? Solve or give some hints.
$\lim_{n\to\infty}\dfrac {C_n^{F_n}}{F_n^{C_n}}$, 
where $C_n=\dfrac {(2n)!}{n!(n+1)!}$ is the n-th Catalan number and $F_n=2^{2^n}+1$ is the n-th Fermat number.

approximate $(n+1)!\simeq n!$ and use sterling approximation
$$L\simeq\lim_{n\rightarrow\infty}\frac{\frac{\sqrt{4\pi n}(\frac{2n}{e})^n}{(\sqrt{2\pi n}(\frac{n}{e})^n)^2}}{2^{2^n}}$$
$$L\simeq\lim_{n\rightarrow\infty}e^{n\ln\left((1/\pi n) (2e/n)\right) -2^nlog2}$$
It appears as n goes to infinity, the upper power part goes i.e. {$\dots$} part of $e^{\dots}$ goes to negative infinity and hence the limit is zero. I took $F_n\simeq2^{2^n}$ and similar approximations.
Since you know $C_n$ grows slower than $F_n$, you can show that for any $a_n$ and $b_n$, if $a_n/b_n$ goes to zero as n goes to infty, $a_n^{b_n}/b_n^{a_n}$ goes to infinity as n goes to infinity.
To show this proceed as above, $$L'=\lim_{n\rightarrow\infty} e^\left(b_n\ln(a_n)-a_n\ln(b_n)\right)$$ since, for any sequence, or function, $\ln(n)$ grows slower than $n$, you can conclude that above limit is infinity.
