Question 1025112
Note that each term is of the form 1/(n*(n+1)) where n ranges from 1 to 219 (*not* 220).


*[tex \large \sum_{n=1}^{219} \frac{1}{n(n+1)}]


In fact, the sum is equal to 1 - (1/220) = 219/220. But that is a completely different problem.