Best solutions to - "How do I show that 2 n /..." - Plainmath

heelallev5

Answered question

2022-08-12

How do I show that

2^{n} / n!

is bounded above?

Answer & Explanation

Riya Cline

Beginner2022-08-13Added 17 answers

0 \leq \frac{2^{n}}{n!} = \frac{2}{1} \frac{2}{2} \frac{2}{3} \dots \frac{2}{n} = 2 \cdot \frac{2}{3} \dots \frac{2}{n} \leq 2 \cdot {(\frac{2}{3})}^{n - 2}

and since

\frac{2}{3} < 1

, we know

{(\frac{2}{3})}^{n - 2} \to 0

as

n \to \infty

For any number replacing 2, say k, eventually n is bigger than k, so this same argument applies.

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