Algebra 2  Geometric Series
posted by Kay on .
I need help evaluating the geometric infinite series.
It's got the double zero thing over the E thing.
00
E(1/2)^ n1
(n=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 
How did you get k + 1?
How would I do
00
E 3(0.4)^n1
n=1
Would that be the same basic thing? Like 1/[13(0.4)]?? 
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/(1a)