- #1
ChelseaL
- 22
- 0
Use the fact that [tex]\frac{1}{k}[/tex] -[tex] \frac{1}{k+1}[/tex] = [tex]\frac{1}{k(k+1)}[/tex] to show that
n
sigma ([tex]\frac{1}{k(k+1)}[/tex]) = 1-[tex] \frac{1}{n+1}[/tex]
r=1
What do I need to do to solve it?
n
sigma ([tex]\frac{1}{k(k+1)}[/tex]) = 1-[tex] \frac{1}{n+1}[/tex]
r=1
What do I need to do to solve it?