Mathematical induction problem: $\frac12\cdot \frac34\cdots\frac{2n-1}{2n}<\frac1{\sqrt{2n}}$

This is a problem that I tried to solve and didn't come up with any ideas .?$$\frac{1}{2}\cdot \frac{3}{4}\cdots\frac{2n-1}{2n}<\frac{1}{\sqrt{2n}}.$$ All I get is $\frac{1}{\sqrt{2n}}\cdot\frac{2n+1}{2n+2}<\frac{1}{\sqrt{2n+2}}$ which evaluates to $1<0$Do you know what to do here ?

Solution 1:

Prove by induction the stronger inequality: for any integer $n\geq 1$, $$\frac{1}{2}\cdot \frac{3}{4}\cdots\frac{2n-1}{2n}<\frac{1}{\sqrt{2n+1}}.$$ The basic step is true: $1/2<1/\sqrt{3}$. Then in the inductive step you will obtain the inequality $$\frac{1}{\sqrt{2n+1}}\cdot\frac{2n+1}{2n+2}<\frac{1}{\sqrt{2n+3}}$$ that is equivalent to $$4n^2+8n+3=(2n+1)(2n+3)<(2n+2)^2=4n^2+8n+4$$ which holds.

Solution 2:

An easier proof without induction:

Let $r$ be a positive integer, whence we have by the AM-GM inequality,

$\dfrac{(2r-1)+(2r+1)}{2}>\sqrt {(2r-1)(2r+1)}$

$\Rightarrow \dfrac{2r}{2r-1}>\sqrt {\dfrac{2r+1}{2r-1}}$

$\Rightarrow \displaystyle\prod_{r=1}^n\left( \frac{2r}{2r-1}\right)>\prod_{r=1}^n\left(\sqrt {\frac{2r+1}{2r-1}}\right)$

$\Rightarrow \dfrac{2.4.6...(2n)}{1.3.5...(2n-1)}>\sqrt {2n+1}$

$\Rightarrow \dfrac{1.3.5...(2n-1)}{2.4.6...(2n)}<\dfrac{1}{\sqrt{2n+1}}<\dfrac{1}{\sqrt {2n}}. $

Solution 3:

Induction is a really efficient way for proving that $$\frac{(2n-1)!!}{(2n)!!} = \frac{(2n)!}{4^n n!^2} = \frac{1}{4^n}\binom{2n}{n}<\frac{1}{\sqrt{2n}}\tag{1}$$ but it isn't the only option. For instance, we may notice that the LHS of $(1)$ can be written as $$\prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)=\sqrt{\prod_{k=1}^{n}\left(1-\frac{1}{k}+\frac{1}{4k^2}\right)} \tag{2}$$ hence: $$ \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)^2 = \frac{1}{4}\color{blue}{\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)}\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)=\frac{1}{4n}\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)\tag{3} $$ since the blue term is a telescopic product. Since $1+z<e^{z}$ for any $z>0$, we also have: $$ \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)^2 < \frac{1}{4n}\exp\color{purple}{\sum_{k=2}^{n}\frac{1}{4k(k-1)}}=\frac{1}{4n}\exp\left(\frac{n-1}{4n}\right)<\frac{e^{1/4}}{4n} \tag{4}$$ since the purple sum is a telescopic sum. By switching to square roots it follows that $$ \frac{(2n-1)!!}{(2n)!!}=\frac{1}{4^n}\binom{2n}{n}< \color{red}{\frac{e^{1/8}}{2\sqrt{n}}}\tag{5}$$ that is a stronger inequality. If we directly compute $\prod_{k=2}^{+\infty}\left(1+\frac{1}{4k(k-1)}\right)=\frac{4}{\pi}$ through Wallis product, with the same approach we may also prove the sharper $$\boxed{ \frac{(2n-1)!!}{(2n)!!}=\frac{1}{4^n}\binom{2n}{n}< \color{red}{\frac{1}{\sqrt{\pi\left(n+\frac{1}{4}\right)}}}.}\tag{6}$$

Solution 4:

This proof true when $\frac{1}{2}\cdot \frac{3}{4}\cdots\frac{2n-1}{2n}<\frac{1}{\sqrt{2n+1}}.$ $$\frac { 1 }{ 2 } \cdot \frac { 3 }{ 4 } \cdots \frac { 2n-1 }{ 2n } \frac { 2n+1 }{ 2n+2 } <\frac { 1 }{ \sqrt { 2n+1 } } .\frac { 2n+1 }{ 2n+2 } =\frac { 1 }{ \sqrt { 2n+3 } } .\frac { \sqrt { 2n+3 } }{ \sqrt { 2n+1 } } \frac { 2n+1 }{ 2n+2 } =\\ =\frac { 1 }{ \sqrt { 2n+3 } } \sqrt { \frac { 4{ n }^{ 2 }+8n+3 }{ 4{ n }^{ 2 }+8n+4 } } <\frac { 1 }{ \sqrt { 2n+3 } } $$