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In the textbook "A classical introduction to modern number theory" 1990 edition, at page 22 they write that

if $n>3$ then $e^{n-1}>2^n$.

I am not sure I see why, I mean if $n>3$ then $e^n /2^n > (e/2)^3$ but the RHS isn't greater than e, right?

Any hints?

Thanks in advance.

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You have asked 6 questions before this but accepted answers for none. Please consider accepting answers for some previous questions if you found them helpful. Thanks. – Srivatsan Sep 23 '11 at 5:01
up vote 4 down vote accepted

Write $2^n$ as $e^{n \ln(2)}$ and use the fact that the exponential function is monotone increasing and $3>4\ln(2)$.

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In a sense, you are correct. $(e/2)^3 \approx 2.510\ldots$, so the inequality seems incorrect as stated.

However, given the context and the notation, I reasonably assume that $n$ is an integer. Therefore, $n > 3 \iff n \geq 4$. Then the inequality becomes correct because $(e/2)^4 \approx 3.412\ldots > 3 > e$.

Alternately, the inequality $e^{n-1} > 2^{n}$ is true iff $$ n > \frac{\log e}{\log e - \log 2} \approx 3.2588\ldots. $$ If $n$ is an integer, this is equivalent to $n \geq 4$ or $n > 3$.

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You want to prove that $(e/2)^n>e$ for $n>3$, that is, for $n\ge 4$. Since $e>2$, $(e/2)^n$ is increasing and it suffices to prove that $(e/2)^4 > 3$, because $3>e$. Using $e>1+1+1/2+1/6=8/3$ gets this done. (I think one point here is to prove it without using too many decimals of $e$.)

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