Proving the upper bound of the ratio of product of odd n numbers to even n numbers?

So question demands a proof that:

Let $$x_n = \frac 12 * \frac34 * \frac56 * ... * \frac{2n-1}{2n}$$

Then show that;

$$x_n \leq \frac 1{\sqrt{3n +1}}$$

So essentially what I have tried to do is use the formulas for the product of first N odd numbers and product of first N even numbers. This gives me that

$$x_n = \frac{^{2n} C_n}{2^{2n}}$$

I have no clue if I have taken the right road. Can any one help me out with this proof ? Thanks

• Have you tried induction? – Servaes Feb 4 at 15:32
• No I have not. I frankly do not know how to use induction for good proof building. Like the condition that required for using induction, what constitutes as proof in this method, etc. – DS112 Feb 4 at 15:32
• There is plenty of literature available on induction. Or have a look at wikipedia. If you prefer to avoid induction, please clarify which methouds you would like to use. – Servaes Feb 4 at 15:34
• @Servaes induction looks like will be troublesome with that $\sqrt{\cdot}$ in the denominator... – gt6989b Feb 4 at 15:34
• @gt6989b That is easily solved by squaring. – Servaes Feb 4 at 15:42

Here's a sketch of a proof by induction; the base case is easy checked as for $$n=1$$ you have $$x_1=\frac{1}{2}\leq\frac{1}{\sqrt{3\cdot1+1}}.$$ Then the induction step; suppose $$x_n\leq\frac{1}{\sqrt{3n+1}}$$ for some $$n\geq1$$. We want to show that this implies $$x_{n+1}\leq\frac{1}{\sqrt{3(n+1)+1}}.$$ By definition of $$x_{n+1}$$ and $$x_n$$, and by the induction hypothesis, we have $$x_{n+1}=\frac{2n+1}{2n+2}x_n\leq\frac{2n+1}{2n+2}\frac{1}{\sqrt{3n+1}},$$ so now it suffices to show that $$\frac{2n+1}{2n+2}\frac{1}{\sqrt{3n+1}}\leq\frac{1}{\sqrt{3(n+1)+1}}.$$ Clearing denominators and squaring (where we use that all terms are positive) shows that this is equivalent to $$(2n+1)^2(3n+4)\leq(2n+2)^2(3n+1),$$ and expanding both sides yields the obviously true statement $$12n^3+28n^2+19n+4\leq12n^3+28n^2+20n+4.$$