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.
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1Solve or tell us what you've done. – Julien Mar 23 '13 at 17:27
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I found some reasons to coclude that $\lim_{n\to\infty}\dfrac {C_n}{F_n}=0$, is that enough for the above mentioned problem? – Mar 23 '13 at 17:32
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Is your question does the limit exist? – Mar 23 '13 at 17:40
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Yes, if it exists I do not need to know its value. – Mar 23 '13 at 17:43
1 Answers
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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.