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$\sum\limits_{x=1}^{99}\frac{1}{x(x+1)}$ = ______.
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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$
Answer:

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