Question

How do you evaluate a infinite summation? Is this right? (click down)

Answer 1

s = a1/(1-r)

0/[1-(1/2)] = 0

Answer 2

no, a1=1/2. because if you plug in i=1, you get 1/2

Answer 3

so the sum is (1/2)/[1-1/2] = 1

Answer 4

the index starts at 0 so the sum should be 2 because \(\large \sum_{i=1}^{\infty}(\frac{1}{2})^t =1\)

Answer 5

thanks everyone