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limitelite

This problem was posted half a year ago by Pierre Mounir on a Facebook group and until now it received no answers. Since most of his problems that I saw were amazing I can bet this one it's worth the time. Wolfram returns the answer to be $2$, which is quite elegant for it's look.

I remembered about it yesterday and gave a try again taking for simplicity $k=1$ (I had no chance with a bigger number). Also my whole idea was to somehow get to a point where I can use $\lim\limits_{f\to 0}\frac{a^f-1}{f}=\ln a$, thus I started as: $$\lim_{n\to \infty} {\frac{5^\frac{n!}{(2n)!}-4^\frac{1}{n!}}{3^\frac{n!}{(2n)!}-2^\frac{1}{n!}}}=\lim_{n\to \infty} \left(\frac{4}{2}\right)^{\frac{1}{n!}}\left(\frac{5^\frac{n!}{(2n)!}}{4^{\frac1{n!}}}-1\right)\left(\frac{3^\frac{n!}{(2n)!}}{2^{\frac1{n!}}}-1\right)^{-1}$$ $$=\lim_{n\to \infty} \underbrace{\sqrt[n!]{2}}_{\to 1}\left(\sqrt[n!]{\frac{5^\frac{1}{(2n)!}}{4}}-1\right)\left(\sqrt[n!]{\frac{3^\frac{1}{(2n)!}}{2}}-1\right)^{-1}$$ Well, yes $\frac{5^\frac{1}{(2n)!}}{4}$ and the other one in the denominator equals to $1$, but still I don't see how to use that limit. Also I tried to take a logarithm on both sides or to use L'hospital, but looks like a dead end.

I would love if someone can spot the trick for solving this limit and land some help.

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    $\begingroup$ Maybe it could be a good idea to use Stirling, just to spot what the limit is? $\endgroup$ – Crostul Apr 20 at 12:23
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    $\begingroup$ If the given expression is $(a-b) /(c-d) $ then $a, b, c, d$ tend to $1$ and hence they can be replaced by their logs. $\endgroup$ – Paramanand Singh Apr 20 at 12:34
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    $\begingroup$ The answer easily comes to be 2 via logs. One just needs to prove that $(n!) ^{2 k} /(2kn)!\to 0$ which is sort of not that difficult. $\endgroup$ – Paramanand Singh Apr 20 at 12:40
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    $\begingroup$ A pedantic answerer could go ahead and use Stirling's formula, since the forbidden usage is misspelled Strirling's. $\endgroup$ – GEdgar Apr 20 at 12:59
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This is an expansion of my comment into an answer.


If the expression under limit is of the form $(A-B) /(C-D) $ then all of $A, B, C, D$ tend to $1$ and we can write $$A-B=B\cdot\frac{\exp(\log A-\log B) - 1}{\log A - \log B} \cdot(\log A - \log B) $$ The first two factors tend to $1$ and hence the numerator can be replaced by $\log A-\log B$. Proceeding in similar manner we see that the expression under limit can be simplified greatly as each term gets replaced by its logarithm.

The resulting expression is $$\frac{a\log 5-2b\log 2}{a\log 3 - b\log 2}$$ where $a=(n!) ^{k} /(2kn)!,b=(n!)^{-k}$ and clearly $a/b=(n!) ^{2k}/(2kn)!$ tends to $0$ so that the desired limit is $2$.


The limiting behavior of $a_n=(n!) ^{2 k} /(2kn)!$ can be concluded via ratio test. We have $$\frac{a_{n+1}}{a_n} =\frac{(n+1)^{2k}}{(2kn+1)\cdots(2kn+2k)}\to\frac{1}{(2k)^{2k}}<1$$ and hence $a_n\to 0$.

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    $\begingroup$ And "$(n!) ^{2k}/(2kn)!$ tends to zero" does not require Stirling's formula. $\endgroup$ – GEdgar Apr 20 at 12:58
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    $\begingroup$ @GEdgar: I did not check that but it appears one can use ratio test or something similar to conclude that it tends to $0$. $\endgroup$ – Paramanand Singh Apr 20 at 12:59
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    $\begingroup$ @GEdgar: finally added a demonstration via ratio test. $\endgroup$ – Paramanand Singh Apr 20 at 13:05
  • $\begingroup$ Uff, I should remove scary from the name since you made it look easy. I didn't realise how useful is to write $\frac{5^\frac{n!}{(2n)!}}{4^{\frac1{n!}}}$ in the $e^{\log }$ form. There I too tried to get into that special limit from, but taking into $\lim\limits_{f(x)\to 0}\frac{a^{f(x)}-1}{f(x)},\,$ the $f(x)$ term as $\log A-\log B$ was clearly what I needed, thanks alot! $\endgroup$ – Number Apr 20 at 13:14
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    $\begingroup$ @Zacky: I am sure it would appear damn scary to many people. $\endgroup$ – Paramanand Singh Apr 20 at 13:15

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