The sum telescopes nicely:
[tex]\displaystyle\sum_{n=1}^{100}a_n=\left(\frac11-\frac12\right)+\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)+\cdots+\left(\frac1{99}-\frac1{100}\right)+\left(\frac1{100}-\frac1{101}\right)[/tex]
[tex]\implies\displaystyle\sum_{n=1}^{100}a_n=1-\frac1{101}=\boxed{\frac{100}{101}}[/tex]
