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

  • $\begingroup$ Have you tried induction? $\endgroup$ – Servaes Feb 4 at 15:32
  • $\begingroup$ 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. $\endgroup$ – DS112 Feb 4 at 15:32
  • $\begingroup$ 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. $\endgroup$ – Servaes Feb 4 at 15:34
  • $\begingroup$ @Servaes induction looks like will be troublesome with that $\sqrt{\cdot}$ in the denominator... $\endgroup$ – gt6989b Feb 4 at 15:34
  • 1
    $\begingroup$ @gt6989b That is easily solved by squaring. $\endgroup$ – 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 $$12n^3+28n^2+19n+4\leq12n^3+28n^2+20n+4.$$

  • 1
    $\begingroup$ Thanks a lot. This resolves my doubt. Also serves as an alarm for me to learn induction based proofs. $\endgroup$ – DS112 Feb 4 at 15:51

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