Under what circumstances does the infinite series 1+(2x/3)+(2x/3)^2+(2x/3)^3+... converge and what is the sum when x=1.2 ?
1 Answer
Sep 6, 2016
The series converges when:
-3/2 < x < 3/2
If
1+(2x/3)+(2x/3)^2+(2x/3)^3+... = 5
Explanation:
This is a geometric series with initial term
A non-zero geometric series with common ratio
In our case, that means:
abs((2x)/3) < 1
Multiplying both sides by
abs(x) < 3/2
which expands to mean:
-3/2 < x < 3/2
The sum of an geometric series with initial term
sum_(n=1)^oo ar^(n-1) = a/(1-r)
So given
sum_(n=1)^oo ar^(n-1) = 1/(1-0.8) = 1/0.2 = 5