ok i have a new problem,
99
E (1/i) - (1/(i+1))
i=3
i simplify it to
E 1
----------
E i^2 + E i
how do i move on, because now i = 3 in the E
Check it out
$\displaystyle \sum_{i=3}^{99}\left(\frac{1}{i}-\frac{1}{i+1}\right)=\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{98}-\frac{1}{99}\right)+\left(\frac{1}{99}-\frac{1}{100}\right)$
So, you can see all of the cancellation that is gonna occur! The only thing that'll be left standing is
$\displaystyle =\frac{1}{3}-\frac{1}{100}$