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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