Question

I need help evaluating the geometric infinite series.

It's got the double zero thing over the E thing.

00
E(-1/2)^ n-1
(n=1)

Answer 1

Substitute:

n = k + 1

Then k runs from zero to infinity

The summand is (-1/2)^k

So, the summation is:

1/[1- (-1/2)] = 2/3

Answer 2

How did you get k + 1?
How would I do
00
E 3(0.4)^n-1
n=1

Would that be the same basic thing? Like 1/[1-3(0.4)]??

Answer 3

You are free to shift your summation variable in any way you like. If you put n = k + 1 then what happens is that the lower limit of the summation of
n = 1 corresponds to k = 0 while the upper limit remains infinity.

In the summand you then replace n by
k + 1 and you then see that the summation is now in the standard form of a sum from zero to infinity of a^k which is 1/(1-a)