The inequality of the series defined by the ratio of the product of odd numbers from 1 to 99 over the product of even numbers from 2 to 100 is proven to be greater than \(\dfrac{1}{10\sqrt{2}}\). The mathematical expression is represented as \(\dfrac {1\times 3 \times 5 \times \ldots \times 99}{2\times 4 \times 6 \times \ldots \times 100} > \dfrac {1}{10\sqrt{2}}\). This conclusion is reached through a series of algebraic manipulations and inequalities, confirming the validity of the statement.
PREREQUISITESMathematicians, students studying advanced algebra, and anyone interested in inequalities and series proofs will benefit from this discussion.
Albert said:prove :
$\dfrac {1\times 3 \times 5\times------\times 99}{2\times 4\times 6\times --\times {100}}>\dfrac {1}{10\sqrt 2}$
very nice !kaliprasad said: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 $\gt\dfrac{1}{10\sqrt{2}}$