Question

pls help

Answer 1

@mathmate

Answer 2

I already know the answer. thx

Answer 3

Good that you have the answer, but here's the process.

Using the same summation as in the previous problem,
\(\large \sum_{k=1}^{n} \frac{1}{k(k+1)}=\sum_{k=1}^{n} (\frac{1}{k}-\frac{1}{k+1})=1-\frac{1}{n+1}=\frac{n}{n+1}\)
using summation of a telescopic series.
Here n=2005, so sum=\(\large \frac{2005}{2006}\)