Reciprocal of a Loooong Sum

Note that

$\sum\limits_{i=1}^{100}\frac{1}{i(i+1)}=\frac{1}{1(2)}+\frac{1}{2(3)}+\cdots+\frac{1}{99(100)}+\frac{1}{100(101)}$

Since

$\frac{1}{i(i+1)}=\frac{1}{i}-\frac{1}{i+1}$

$\frac{1}{1(2)}+\frac{1}{2(3)}+\cdots+\frac{1}{99(100)}+\frac{1}{100(101)}$

$=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}\cdots+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}$

$=\frac{1}{1}-\frac{1}{101}$

$=\frac{100}{101}$

Hence the desired answer is $\frac{1}{\frac{100}{101}}=\frac{101}{100}=\boxed{1.01}$ .

By using this fact,

$\frac{1}{i(i+1)} = \frac{1}{i}- \frac{1}{i+1}$

we could compute that the denominator's value would become $1 - \frac{1}{101} = \frac{100}{101}$ .

Which lead us to $\frac{1}{(\frac{100}{101})} = \frac{101}{100} = 1.01$ as the solution of the problem.