Add and Subtract x in Numerator
i.e $\frac{1}{x(x+1)} = \frac{x+1-x}{x(x+1)} = \frac{x+1}{x(x+1)} -\frac{x}{x(x+1)}= \frac{1}{x} - \frac{1}{x+1}$
Then,
$\sum_{x=1}^{99}(\frac{1}{x} -\frac{1}{x+1}) \: = \: \frac{1}{1}-\frac{1}{2}\:\: +\frac{1}{2}-\frac{1}{3}\: \: +\frac{1}{3}-\frac{1}{4}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot -\frac{1}{99}+\frac{1}{99}-\frac{1}{100}$
$\therefore$
$1-\frac{1}{100}\: \: \; = \frac{99}{100}$$1-\frac{1}{100}\: \: \; = \frac{99}{100} \: \: = 0.99$