- #1
Lance WIlliam
- 47
- 0
[tex]\sum[/tex] as n=0(theres a infinity above that sigma) , 3(1/11)^n
I thought it would just be 3/11 and converge due to the geometric test but its not...to find the sum would I just start at 0 and put numbers in for "n"...
I thought it would just be 3/11 and converge due to the geometric test but its not...to find the sum would I just start at 0 and put numbers in for "n"...