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If we know that $\frac{2^n}{n!}>0$ for every $n\in \mathbb{N}$ and $$\frac{2^n}{n!}=\frac{2}{1}\frac{2}{2}...\frac{2}{n}$$ how to bound this sequence above?

Martin Sleziak
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user300045
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    The Stirling's approximation tells that $n! \sim \sqrt{2\pi n}(\frac{n}e)^n$ – Mythomorphic Jun 10 '15 at 10:33
  • If you don't know Stirling's approximation you could use Hopital's rule (this is a $\frac {\infty}{\infty}$ form) but to use it you have to know Gamma function and how to derive it... – AlienRem Jun 10 '15 at 10:38
  • Whenever you have a problem which involves factorials, think immediately Stirling approximation. It always makes life simple. – Claude Leibovici Jun 10 '15 at 10:44

4 Answers4

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For $n \ge 4$ you have $$\frac{2}{1} \frac{2}{2} \frac{2}{3} \cdots \frac{2}{n} \le 2 \cdot 1 \cdot 1 \cdot \dots 1 \cdot \frac{2}{n} = \frac{4}{n}$$

Crostul
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You cannot know that $\lim_{n\to +\infty}\frac{2^n}{n!}\color{red}{>}0$ since the limit is exactly zero.

That follows from the trivial inequality: $$ \forall n\geq 3,\qquad \frac{2^n}{n!}\leq \frac{2^n}{2\cdot 3^{n-2}}=\frac{9}{2}\cdot\left(\frac{2}{3}\right)^n\xrightarrow[n\to +\infty]{}0.$$

Jack D'Aurizio
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$$a_n:=\frac{2^n}{n!}\implies\frac{a_{n+1}}{a_n}=\frac{2^{n+1}}{(n+1)!}\frac{n!}{2^n}=\frac2{n+1}\xrightarrow[n\to\infty]{}0$$

so by the ratio test (d'Alembert's), we get that the series

$$\sum_{n=1}^\infty\frac{2^n}{n!}\;\;\;\text{converges}\;\;\;\implies\;\;\;\lim_{n\to 0}\frac{2^n}{n!}=0$$

Timbuc
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Hint: Stirling's approximation

Atvin
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