calculus
posted by Jace on .
I'm having trouble with a geometric series problem.
Determine if
the infinite summation of
(3)^(n1)/4^n
converges or diverges. If it converges, find the sum.
So the answer says that
sigma 3^(n1)/4^n
= 1/4 * sigma (3/4)^(n1)
How did they factor out the 1/4? My algebra is very weak :(

the first part:
4^n can be written as
4(4)^(n1) = (1/4) (4)^(n1)
then (3)^(n1)/4^n
= (1/4) (3)^(n1) / 4^(n1)
= (1/4) (3/4)^(n1)
no now:
∑ (1/4) (3/4)^(n1)
= (1/4)(3/4)^0 + (1/4)(3/4)^1 + (1/4)(3/4)^2 + ...
which is an infinitite geometric series , which clearly converges.
with a = (1/4) and r = 3/4
sum∞ = a/(1r) = (/4)/(1(3/4)
= (1/4)/(7/4)
= 1/7