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.

Let $R>2$ be a real number. Then, for any $n\geq 1$, it holds that

$$ \frac{ \exp(n R)+ 1}{\exp( n R) - \exp(\pi R/2)} \leq \exp( R/n^2).$$

How do I prove this?

share|improve this question
2  
For $n=1$ the inequality: $$e^R+1\leq e^R(e^R-e^{\pi R/2})$$ is false, because $e^{(1+\pi/2)R}\geq e^{2R}-e^R-1$ for $R\geq 0$. –  Pacciu Dec 5 '11 at 16:19

1 Answer 1

The inequality as stated: $$ \frac{ \exp(n R)+ 1}{\exp( n R) - \exp(\pi R/2)} \leq \exp( R/n^2) $$ can not be true for all $R>2$ and $n \geq 1$. Indeed, choose $n=2$, then for all $2 < r < r_\ast$, the inequality does not hold, where $r_\ast$ is the unique positive root of $$ e^{2 r}-e^{9 r/4}+\exp\left({\frac{\pi r}{2}+\frac{r}{4}}\right)+1 $$ approximately equal $r_\ast \approx 2.118953$.

enter image description here

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.