Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I would like to show that $\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\leq \frac{1}{\sqrt{3n+1}}$ holds for all natural numbers. I got stuck here:

$\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\cdot\frac{2n+1}{2n+2}\leq \frac{1}{\sqrt{3n+1}}\frac{2n+1}{2n+2}.$

I would appreciate your help.

share|cite|improve this question
Show that $\frac1{\sqrt{3n+1}} \frac{2n+1}{2n+2} \leq \frac1{\sqrt{3(n+1)+1}}$. – user17762 Mar 13 '12 at 18:54
up vote 6 down vote accepted

You want to show that $$ \frac1{\sqrt{3n+1}}\;\frac{2n+1}{2n+2}\leq\;\frac1{\sqrt{3n+4}}. $$ Since everything is positive, this inequality is the same as $$ (3n+4)(2n+1)^2\leq (3n+1)(2n+2)^2. $$ After expanding and cancelling the $n^3$ terms we get $$ 12n^2+19n\leq 24n^2+20n. $$ This inequality holds trivially for any $n\in\mathbb{N}$, and now you can retrace the steps back to your desired inequality.

share|cite|improve this answer

Your Answer


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.