If we substitute the first three lines into the series of fractions we will get

$\displaystyle 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...$

Looking at this number pattern, we can deduce that the last two fraction will be $\displaystyle +\frac{1}{{98}}-\frac{1}{{99}}+\frac{1}{{99}}-\frac{1}{{100}}$

Therefore,

$\displaystyle 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{{98}}-\frac{1}{{99}}+\frac{1}{{99}}-\frac{1}{{100}}$

$\displaystyle -\frac{1}{2}+\frac{1}{2}$ will cancel each other, so will$\displaystyle -\frac{1}{3}+\frac{1}{3}$,

we can safely assume that all the fraction in the series will cancel out  its neighbor

Thus the remainder will be $\displaystyle 1-\frac{1}{{100}}=\frac{{99}}{{100}}$

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