I need help evaluating the geometric infinite series.
It's got the double zero thing over the E thing.
Algebra 2 - Geometric Series - Count Iblis, Wednesday, April 29, 2009 at 1:15pm
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
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)