How can I prove without using a calculator that $\frac{1}{e} > \frac{\ln \pi}{\pi}$?

Question

Without using a calculator.

I can see that $\ln \pi$ is close to $1$ but a little bit greater... Since $e$ is less than $\pi$, $\frac{1}{e}$ has to be a larger number. I don't understand how someone could know $\ln \pi$ though without using a calculator?

I also then have to prove why $e^\pi > \pi^e$.

http://math.stackexchange.com/questions/7892/comparing-pie-and-e-pi — Surb, Dec 04 '15 at 15:01
@Surb: Sorry to ask this here. how could you find the related question to this? :) In most of the times I am unable to do so by searching!? Is there any other way or a special trick? :) — Hosein Rahnama, Dec 04 '15 at 15:20
@H.R. No... I think searching methods have been extensively discussed in meta. But that's what I did and then I looked at the source of the duplicate. — Surb, Dec 04 '15 at 15:22
@Surb: Many Thanks, I thought that we can just search words not equations! :) — Hosein Rahnama, Dec 04 '15 at 15:25
@H.R. Looking at the searching output, it seems that I found the question because of the "greater than" and not of the $e^{\pi}$... — Surb, Dec 04 '15 at 15:31

score 1 · Answer 1 · answered Dec 04 '15 at 16:28

The function $f(x)=\frac{\ln x}{x}$ goes from $-\infty$ to $0$ and its only maximun is given by $f(e)=\frac 1e$ (because $f'(x)=\frac{1-\ln x}{x^2}$) so $f$ is increasing on the interval $]0,e[$ and decreasing on $]e,\infty[$.

On the other hand $\frac{1}{e} > \frac{\ln \pi}{\pi}\iff \frac{\ln e}{e} > \frac{\ln \pi}{\pi}$ and since $e<\pi$ we finish because of the interval where $f$ is decreasing.

How can I prove without using a calculator that $\frac{1}{e} > \frac{\ln \pi}{\pi}$?

1 Answers1