Is my proof using squeeze theorem correct?

Asked Apr 19 '21 at 17:18

Active Apr 19 '21 at 17:23

Viewed 40 times

I have to prove using squeeze theorem that limit of $(1.3.5...(2n-1))/(2.4.6...2n)$ tends to 0 as n tends to infinity.

My attempt:

$1/(2.4.6...2n)\leq u_n\leq ((1.3.5...(2n-1))/(2.2.2...2n)$, where $u_n$ is the given sequence.

Thus we have, $1/(2.4.6...2n)\leq u_n\leq ((1.3.5...(2n-1))/2^{n}(n)$

Taking limit on both the sides, we get,

$0\leq lim u_n\leq 0$

Is this correct??

Thanks in advance.

edited Apr 19 '21 at 17:23

asked Apr 19 '21 at 17:18

Natasha J

have you heard of the Stirling's approximation? – Fei Cao Apr 19 '21 at 17:22
No, I haven't heard of it! – Natasha J Apr 19 '21 at 17:23
Are you asking the limit of$ (2n-1)/2n$? – Vivaan Daga Apr 19 '21 at 17:26
No I am asking to prove the limit of (1.3.5...(2n-1))/(2.4.6...2n) – Natasha J Apr 19 '21 at 17:28
2

Why do you think the limit of the right-hand side is 0? – Dan Velleman Apr 19 '21 at 17:38
Then what should I take my right hand side as? – Natasha J Apr 19 '21 at 17:50
This is a tricky problem. Is it homework for a class? Were you given a hint? Was any similar problem discussed in class? – Dan Velleman Apr 19 '21 at 18:03
Hint: Can you discover $\alpha \in (0,1)$ such that $u_n \leq n^{-\alpha}$ for all $n\in\mathbb{N}$? Have in mind that you will be proving $u_n \leq n^{-\alpha}$ for all $n\in\mathbb{N}$ by Mathematical Induction. – wdacda Apr 19 '21 at 20:12

0 Answers0