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Could I please have a hint for finding the following limit?$$\lim_{n\to\infty}\frac{5n^n}{3n!+3^n}$$

Martin Sleziak
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Caleb Owusu-Yianoma
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    HINT: try to consider $\lim_n \frac{3n!+3^n}{5n^n}$ – Crostul May 19 '15 at 10:43
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    Maybe try to apply Stirling's approximation. – rwols May 19 '15 at 10:43
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    Consider $\frac{6\cdot 6\cdot 6\cdot 6\cdot 6\cdot 6}{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6}$. – Nikolaj-K May 19 '15 at 10:49
  • @Crostul: I am working on your suggestion and have reached a point where I must calculate $\lim_{n\to\infty}\frac{n^n}{(n+1)^{n+1}}$. How could I go about this in a rigorous way, please? – Caleb Owusu-Yianoma May 19 '15 at 10:57
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    @CKKOY the limit you want you can get if you rewrite the potencies as exponentials and you will get that the limit is 1/n (so 0). exp(x*(log(x)-log(x+1))-log(x+1)) -> exp(-log(x)) -> 1/x – Wolfgang Brehm May 19 '15 at 11:15
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    Some of these older posts might help you: http://math.stackexchange.com/questions/535226/limit-of-the-sequence-nn-n, http://math.stackexchange.com/questions/61713/whats-the-limit-of-the-sequence-lim-n-rightarrow-infty-fracnnn, http://math.stackexchange.com/questions/397866/limits-of-sequences-exponential-and-factorial – Martin Sleziak May 19 '15 at 11:20
  • Especially this approach seems easy: http://math.stackexchange.com/questions/61713/whats-the-limit-of-the-sequence-lim-n-rightarrow-infty-fracnnn/61741#61741 – Martin Sleziak May 19 '15 at 11:23

3 Answers3

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Stirlings approximation of the factorial: $$ \sqrt{2\,\pi\,n}\left(\frac{n}{e}\right)^n $$ 5 and 3 are constant factors so put them before the limit. Now we have to concern ourselves with: $$ \frac{5}{3}\lim_{n\rightarrow\infty}\dfrac{n^n}{\sqrt{2\,\pi\,n}\left(\frac{n}{e}\right)^n+3^n} = \frac{5}{3\,\sqrt{2\,\pi}}\lim_{n\rightarrow\infty}\dfrac{n^n}{\sqrt{n}\frac{n^n}{e^n}+3^n} = \frac{5}{3\,\sqrt{2\,\pi}}\lim_{n\rightarrow\infty}\dfrac{1}{\dfrac{\sqrt{n}}{e^n}+\left(\frac{3}{n}\right)^n}\\ = \frac{5}{3\,\sqrt{2\,\pi}}\lim_{n\rightarrow\infty}{\dfrac{e^n}{\sqrt{n}}} $$

Wolfgang Brehm
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I think the limit diverges as consider the inverse of limit i.e $$\lim_{n\to\infty} \frac{3n! + 3^n}{5n^n}$$ $$\lim_{n\to\infty} \frac{3}{5}*\frac{3^n}{n^n}$$ tends to zero as 3/n tends to zero as n tends to infinity. Also $$\lim_{n\to\infty} \frac{3}{5}*\frac{n!}{n^n}$$ also tends to zero as $$ e\left(\frac{n}{e}\right)^n \leqslant n! \leqslant ne\left(\frac{n}{e}\right)^n$$ and using sandwich theorem on this we get the total limit tends to zero so the actual limit given diverges.

Chappers
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user313732
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Use equivalents:

First $3n!+3^n=3n!+o(n!)$ so: $$ 3n!+3^n\sim_\infty3n!\sim_\infty 3\sqrt{2\pi n}\Bigl(\frac n{\mathrm e}\Bigr)^n$$ and $$\frac{5n^n}{3n!+3^n}\sim_\infty\frac{\mathrm e^n}{3\sqrt{2\pi n}}\xrightarrow[n\to\infty]{}\infty.$$

Bernard
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