I managed to prove the sequence is monotonic and bounded which means it is convergent. Though, I couldn't evaluate the limit itself. I tried to use a comparation test with no results. Anyway, the result is 0. Any help will bi appreciated.
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There is a typo in the title of the question, please fix it. – YiFan Tey Oct 25 '18 at 21:38
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2Possible duplicate of Limit of $2^{n} n!/n^{n}$ as $n \to \infty$ – Jyrki Lahtonen Oct 26 '18 at 04:31
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HINT
By ratio test
$$\frac{\frac{2^{n+1}(n+1)!}{(n+1)^{n+1}}}{\frac{2^{n}(n)!}{n^{n}}}=2\left(\frac n {n+1}\right)^n=\frac2{\left(1+\frac 1 {n}\right)^n}$$
user
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Hint: use equivalence and Stirling's formula: $$n!\sim_{\infty}\sqrt{2\pi n}\Bigl(\frac n{\mathrm e}\Bigr)^n,$$so $$\frac{2^{n}n!}{n^n}\sim_{\infty}\sqrt{2\pi n}\Bigl(\frac 2{\mathrm e}\Bigr)^n$$ Can you proceed from here?
Bernard
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Bernard, thank you for the hint. I think the next step is to convert the form: $0\infty$ to indeterminate expression: $\frac{0}{0}$ and then to apply l'Hospital's rule,right? – darko sucevic Oct 26 '18 at 11:42
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Not necessary: it only uses a highschool result (if $\alpha>0$ and $,0<a<1,:x^\alpha a^x\to 0$). – Bernard Oct 26 '18 at 11:46