Proving the inequality $\frac12\frac34....\frac{2n-1}{2n}<\frac1{\sqrt{2n+1}}$ How to show this inequality:
$\dfrac{1}{2}\dfrac{3}{4}....\dfrac{2n-1}{2n}<\dfrac{1}{\sqrt{2n+1}}$
Using induction the inequality is verified for $n=1$
now assume that that the inequality holds for $n$,to show it for $n+1$
Then
$\dfrac{1}{2}.\dfrac{3}{4}....\dfrac{2n-1}{2n}\dfrac{2n+1}{2(n+1)}<\dfrac{1}{\sqrt{2n+1}}.\dfrac{2n+1}{2(n+1)}=\dfrac{\sqrt{2n+1}}{2(n+1)}<\dfrac{\sqrt{2(n+1)}}{2(n+1)}=\dfrac{1}{\sqrt{2(n+1)}}$
but I have to make it less than $\dfrac{1}{\sqrt{2n+3}}$ which is not coming.Any help
 A: It's the same proving 
$\enspace\Bigl(\dfrac{1}{2}\dfrac{3}{4}\dots\dfrac{2n-1}{2n}\Bigr)^2<\dfrac{1}{2n+1}$.
Now since $\dfrac ab <\dfrac{a+1}{b+1}$ if and only if  $\dfrac ab <1 $, we can write:
\begin{align*}
\Bigl(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdots\dfrac{2n-1}{2n}\Bigr)^2&=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{3}{4}\dots\dfrac{2n-1}{2n}\cdot\dfrac{2n-1}{2n}\\
&<\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\dots\dfrac{2n-1}{2n}\cdot\dfrac{2n}{2n+1}=\dfrac{1}{2n+1}.
\end{align*}
A: Another posibility is to use the fact that $$\frac{k-1}{k}<\frac{k}{k+1} $$ for any k greater or equal to 1. If you write this inequality for k=3,5,...,2n-1 and multiply them, the answer should occur. 
A: You reached the fact that:
$\dfrac{1}{2}.\dfrac{3}{4}....\dfrac{2n-1}{2n}\dfrac{2n+1}{2(n+1)}<\dfrac{1}{\sqrt{2n+1}}.\dfrac{2n+1}{2(n+1)}=\dfrac{\sqrt{2n+1}}{2(n+1)}$
So it's enaugh to prove that $\dfrac{\sqrt{2n+1}}{2(n+1)}\le\dfrac{1}{\sqrt{2n+3}}$ which is equivalent to: $$\sqrt{(2n+1)(2n+3)}=\sqrt{4n^2+8n+3}\le 2(n+1)$$
By squaring both sides, you can prove that it's correct.
A: Asuume this is true for $n=k$ i.e. $$\dfrac{1}{2}\dfrac{3}{4}....\dfrac{2k-1}{2k}<\dfrac{1}{\sqrt{2k+1}}$$ then you have to show that $$\dfrac{1}{2}\dfrac{3}{4}....\dfrac{2k+1}{2k+2}<\dfrac{1}{\sqrt{2k+3}}$$. But notice that $$\dfrac{1}{2}\dfrac{3}{4}....\dfrac{2k+1}{2k+2}<\dfrac{1}{\sqrt{2k+1}}\cdot \dfrac{2k+1}{2k+2}$$ but $$\dfrac{1}{\sqrt{2k+1}}\cdot \dfrac{2k+1}{2k+2}=\frac{\sqrt{2k+1}}{2k+2}<\frac{1}{\sqrt{2k+3}}$$ The last inequality you can check by cross multiplying and simplifying terms. Thus result is true by induction.
