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

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.
