MHB Is the Inequality of Series Proven with 1 to 99 and 2 to 100?

[Albert1]

prove :
$\dfrac {1\times 3 \times 5\times------\times 99}{2\times 4\times 6\times --\times {100}}>\dfrac {1}{10\sqrt 2}$

 

[kaliprasad]

prove :
$\dfrac {1\times 3 \times 5\times------\times 99}{2\times 4\times 6\times --\times {100}}>\dfrac {1}{10\sqrt 2}$

Spoiler

Let

$A= \frac{3}{4}*\frac{5}{6}\cdots\frac{99}{100}$
$B = \frac{2}{3}*\frac{4}{5}\cdots\frac{98}{99}$
In the above products each term of A is > B pairwise so A > B

$A * B = \frac{2}{100}$ and hence $A \gt\sqrt{(\frac{2}{100})}$

hence $ A \gt\frac{\sqrt{2}}{10}$

hence given expression which is $\frac{A}{ 2}$

so $\gt\dfrac{1}{10\sqrt{2}}$

 

[Albert1]

Spoiler

Let

$A= \frac{3}{4}*\frac{5}{6}\cdots\frac{99}{100}$
$B = \frac{2}{3}*\frac{4}{5}\cdots\frac{98}{99}$
In the above products each term of A is > B pairwise so A > B

$A * B = \frac{2}{100}$ and hence $A \gt\sqrt{(\frac{2}{100})}$

hence $ A \gt\frac{\sqrt{2}}{10}$

hence given expression which is $\frac{A}{ 2}$

so $\gt\dfrac{1}{10\sqrt{2}}$

very nice !