MHB Prove 1000th Progression of an Algebraic Series is Less than 1/79

[anemone]

Prove $\dfrac{1}{3}\cdot\dfrac{4}{6}\cdot\dfrac{7}{9}\cdots\dfrac{1000}{1002}\lt \dfrac{1}{79}$

 

[anemone]

Prove $\dfrac{1}{3}\cdot\dfrac{4}{6}\cdot\dfrac{7}{9}\cdots\dfrac{1000}{1002}\lt \dfrac{1}{79}$

Subtle hint:

Spoiler

Use two other product expressions to make comparison...

 

[anemone]

Prove $\dfrac{1}{3}\cdot\dfrac{4}{6}\cdot\dfrac{7}{9}\cdots\dfrac{1000}{1002}\lt \dfrac{1}{79}$

Solution of other:

Spoiler

Let

$X=\dfrac{1}{3}\cdot\dfrac{4}{6}\cdot\dfrac{7}{9}\cdots\dfrac{1000}{1002}$

$Y=\dfrac{2}{4}\cdot\dfrac{5}{7}\cdot\dfrac{8}{10}\cdots\dfrac{1001}{1003}$

$Z=\dfrac{3}{5}\cdot\dfrac{6}{8}\cdot\dfrac{9}{11}\cdots\dfrac{1002}{1004}$

Note that $X\lt Y\lt Z$, therefore we get $X^3\lt XYZ=\dfrac{1\cdot 2}{1003\cdot 1004}=\dfrac{1}{503506}\lt \dfrac{1}{493039(=79^3)}$.

Thus we have proved that $X=\dfrac{1}{3}\cdot\dfrac{4}{6}\cdot\dfrac{7}{9}\cdots\dfrac{1000}{1002}\lt \dfrac{1}{79}$.