Prove $\frac{1}{2}<S<1$ in Sequence Challenge

[anemone]

Given that $S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4} + ...+\dfrac{1}{99}-\dfrac{1}{100}$.

Prove that $\dfrac{1}{2}<S<1$.

 

[Nono713]

Spoiler

$$S = \left ( 1 - \frac{1}{2} \right ) + \left ( \frac{1}{3} - \frac{1}{4} \right ) + \cdots + \left ( \frac{1}{99} - \frac{1}{100} \right ) > \left ( 1 - \frac{1}{2} \right ) = \frac{1}{2}$$
$$S = 1 - \left ( \frac{1}{2} - \frac{1}{3} \right ) - \left ( \frac{1}{4} - \frac{1}{5} \right ) - \cdots - \frac{1}{100} < 1$$
$\blacksquare$

 

[anemone]

Good job, Bacterius! And thanks for participating!