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)^ n-1

(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)^n-1

n=1

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

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