2

Here is the original question:

Let $$x_n=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots\frac{2n - 1}{2n}$$ Then show that $$x_n\leq\frac{1}{\sqrt{3n+1}} \text{, for all } n = 1, 2, 3, \ldots \text{ .}$$

It is trivial to show that the statement is true for $n=1$.

I assumed it is true for some arbitrary positive integer $k$. Hence, we have

$$x_k\leq\frac{1}{\sqrt{3k+1}}$$

Now, after proper grouping, we arrive at the expression for $x_{k+1}$:

$$ x_{k+1}=x_k\cdot\frac{2k+1}{2k+2} $$ which implies that $$ x_{k+1}\leq \frac{2k+1}{2k+2}\cdot\frac{1}{\sqrt{3k+1}} $$

From here, ho do I arrive at the fact that $$ x_{k+1}\leq \frac{1}{\sqrt{3(k+1)+1}}=\frac{1}{\sqrt{3k+4}}? $$

Hungry Blue Dev
  • 1,708
  • You only need to prove $\frac{2k+1}{2k+2}\cdot \frac{1}{\sqrt{3k+1}} \leqslant \frac{1}{\sqrt{3k+4}}$. – ntt Apr 13 '17 at 20:13
  • Related but possibly not useful: $$x_k=\frac{1}{4^k}{2k\choose k}$$ –  Apr 13 '17 at 20:17
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    This has been asked before: https://math.stackexchange.com/questions/2136612/prove-that-frac12-cdot-frac34-cdots-frac2n-12n-leq-frac1-s?noredirect=1&lq=1, https://math.stackexchange.com/questions/119773/how-does-one-prove-that-frac12-cdot-frac34-cdots-frac2n-12n-leq?noredirect=1&lq=1. – StubbornAtom Apr 14 '17 at 13:06

2 Answers2

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Hint:

Show that $$\frac{(3k+1)(2k+2)^2}{(2k+1)^2}>3k+4$$

via polynomial division.

Thomas Andrews
  • 177,126
1

Say this inequality is correct for $x_n$. So we can say that: $x_n \leq \dfrac{1}{\sqrt{3n+1}}$ $x_{n+1} = x_n * \dfrac{2n+1}{2n+2} \leq \dfrac{1}{\sqrt{3n+1}}*\dfrac{2n+1}{2n+2}$ So know if we prove the statement below, the inequality is proved: $\dfrac{1}{\sqrt{3n+1}}*\dfrac{2n+1}{2n+2} \leq \dfrac{1}{\sqrt{3n+4}}$ In order to do that, we assume that this statement is correct and then use it to derive an obvious statement. Then we can claim that the initial statement that we had, is correct. $\dfrac{1}{\sqrt{3n+1}}*\dfrac{2n+1}{2n+2} \leq \dfrac{1}{\sqrt{3n+4}} => \dfrac{2n+1}{2n+2} \leq \sqrt{\dfrac{3n+1}{3n+4}} => \dfrac{4n^2+4n+1}{4n^2+8n+4}\leq\dfrac{3n+1}{3n+4}=>$

$12n^3+28n^2+19n+4\leq12n^3+28n^2+20n+4=>n\geq0$

Which is an obvious statement which means for any $n\geq 0$ this statement is correct. So the inequality is proved.

Soroush khoubyarian
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  • I like it. Working backward might be useful in a stressful environment, like during an exam. – Hungry Blue Dev Apr 13 '17 at 20:24
  • Of course all of the things you do must be reversible, for example if you say $y=\sqrt{x}$ and then say $y^2 = x$, you have to notice at the end that $y$ must be a positive number. – Soroush khoubyarian Apr 13 '17 at 20:26