Show by induction that (2n-1) /(2n) <= 1/(3n+1) for all n

[Perrault]

Show by induction that (2n-1)!/(2n)! <= 1/(3n+1) for all n

Homework Statement

Show by induction that $(\frac{1\cdot3\cdot5\cdot\cdot\cdot(2n-1)}{2\cdot4\cdot6\cdot\cdot\cdot2n})^2$ $\leq$$\frac{1}{3n+1}$ for n=1,2,3, ...

The Attempt at a Solution

It might not really be relevant, but I've attached my work. I tried to subprove something by induction to help my principal induction, but I got something that was harder than the original.

P.S. I tried using the Physics forums android app, but I couldn't figure out how to post.

 

[genericusrnme]

For n=1,2,3 you have 1/2 <= 1/4, 3/8 <= 1/7 and 5/16 <= 1/10 respectively
are you sure that's the problem you're trying to solve?

 

[Perrault]

Thanks for your response. By "n=1,2,3,...", the problem statement means the natural numbers. Sorry for the confusion.
Also, induction is mandatory.

 

[genericusrnme]

I understand that, but the statement isn't true for all natural numbers since it isn't true for the first at least 10 of them

 

[Perrault]

Of course, I missed something essential! The whole left side is squared. I edited the original post. Thanks for pointing that out.

 

[brmath]

But it still isn't true that $(3/8)^2 < 1/7$

 

[Dick]

But it still isn't true that $(3/8)^2 < 1/7$

(3/8)^2 is less than 1/7.

 

[dirk_mec1]

Yiu'll need to prove that: $$\left( \frac{2n+1}{2n+2} \right) ^2 \leq \frac{3n+1}{3n+4}$$.

The easiest way to do this is to write out the cross product.

 

[brmath]

(3/8)^2 is less than 1/7.

Yes, of course; I managed to square 8 and get 16.

 

[Dick]

Yiu'll need to prove that: $$\left( \frac{2n+1}{2n+2} \right) ^2 \leq \frac{3n+1}{3n+4}$$.

The easiest way to do this is to write out the cross product.

In case it's not obvious, dirk_mec1's idea it that you can express $a_{n+1} \le b_{n+1}$ as $a_{n} \frac{a_{n+1}}{a_n} \le b_{n} \frac{b_{n+1}}{b_n}$. You know $a_n \le b_n$ by your induction hypothesis. If the ratios follow the same pattern you are done.

 

[dirk_mec1]

Indeed, that is exactly my idea and I've written it out to see it leading to succes :).

 

[Perrault]

Awesome guys, it worked! Thanks!