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We know,

$$ e^x-1 = \sum_{i=1}^{\infty} \frac{x^i}{i!}$$

Assume $x>0$ and use am-gm on the equation:

$$ e^x -1\geq \lim_{n \to \infty}n\left( \frac{x^{ \frac{n(n+1)}{2} } }{\prod_{i=1}^n (i!)}\right)^{1/n}$$

What would be the lower bound on RHS for a given $x$? Would it be possible to evaluate the limit in terms of $n$?

p.s: I came up with this myself, I am not sure how to start since we have product of factorial on the right side but for those who want to try my question, Stirling's approximation may be helpful

VIVID
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tryst with freedom
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1 Answers1

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From the known asymptotics of the factorial and the Barnes $G$-function, we have $$ \prod\limits_{i = 1}^n {i!} = n!G(n + 1) = Cn^{n^2 /2 + n} e^{ - 3n^2 /4 - n} (2\pi )^{(n + 1)/2} n^{5/12} \left( {1 + \mathcal{O}\!\left( {\frac{1}{n}} \right)} \right) $$ with a positive constant $C$. Thus $$ \left( {\prod\limits_{i = 1}^n {i!} } \right)^{1/n} = \frac{{\sqrt {2\pi } }}{e}n^{n/2 + 1} e^{ - 3n/4} \left( {1 + \mathcal{O}\!\left( {\frac{{\log n}}{n}} \right)} \right). $$ Accordingly, $$ n\left( \frac{x^{ \frac{n(n+1)}{2} } }{\prod_{i=1}^n i!}\right)^{1/n} = e\sqrt {\frac{x}{{2\pi }}} \left( {e^{3/2} \frac{x}{n}} \right)^{n/2} \left( {1 + \mathcal{O}\!\left( {\frac{{\log n}}{n}} \right)} \right) $$ as $n\to +\infty$ with any $x>0$. This shows that the limit is $0$ and the way it approaches this limit.

Gary
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    Nice answer! So the OP has found a rather complicated way to show $e^x\geq 1$ for $x\geq 0$. – MaoWao Jul 21 '21 at 11:13