Finding limits to a series

Wouldn’t it be nice if we had ….

$$\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times …… \times \frac{99}{100}$$

So, compare some form of S with the above.

What form?

S currently has 50 fractions, whereas the above convenient expression has 99 fractions. So the required form of S also needs to have 99 fractions for easy comparison …. or else make both expressions have 100 fractions, such that the comparison is easy … and note it is just a comparison, which is smaller, which is greater.

A. \(S < \frac{1}{10}\)

S² = \(\frac{1}{2} \times \frac{1}{2} \times \frac{3}{4} \times \frac{3}{4} \times \frac{5}{6} \times \frac{5}{6} \times …… \times \frac{99}{100} \times \frac{99}{100} \)

And say, T = \(\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times \frac{5}{6} \times \frac{6}{7} \times …… \times \frac{99}{100} \times \frac{100}{101} \)

Each of the above expressions have 100 fractions. Of which 50 fractions are the same. Compare each correspondingly placed fraction between the two expressions and decide which which inequality will be replace the ? in

S²  ?  T

Since,

1/2 < 2/3,

3/4 < 4/5,

5/6 < 6/7,

…. and so on right till

99/100 < 100/101

hence we will have

S²  <  T

And T can very easily be reduced to \(\frac{1}{101}\)

Thus, \(S^2 < \frac{1}{101}\)

Being positive quantities, we have \(S < \frac{1}{\sqrt{101}}\)

And, \(\sqrt{101} > \sqrt{100} \implies \frac{1}{\sqrt{101}} < \frac{1}{\sqrt{100}}\)

\(S < \frac{1}{\sqrt{101}} < \frac{1}{\sqrt{100}}\)