Posted by Kay on Wednesday, April 29, 2009 at 1:00pm.
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)

Algebra 2  Geometric Series  Count Iblis, Wednesday, April 29, 2009 at 1:15pm
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

Algebra 2  Geometric Series  Kay, Wednesday, April 29, 2009 at 1:44pm
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)]??

Algebra 2  Geometric Series  Count Iblis, Wednesday, April 29, 2009 at 5:33pm
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)
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