Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
7  
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

3 Answers 3

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 $3>4\ln(2)$.

share|improve this answer

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 $n > 3$.

share|improve this answer

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 $e$.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.