2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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 $$12n^3+28n^2+19n+4\leq12n^3+28n^2+20n+4.$$

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.