# Simple Geometric series

1. Mar 6, 2009

### fball558

1. The problem statement, all variables and given/known data
actually got two questions but both are related so put them in the same place
Determine whether the geometric series is convergent or divergent. If it is convergent, find its sum.
Inf
1.) E 6(0.9)^(n-1)
n=1

Inf (-3)^(n-1)
2.) E ---------------
n=1 4^(n)

the (E) is the sigma sign asking for sum
the Inf is infinity and n = 1 is inital starting point
they want us to evaluate the series sum from 1 to infinity

3. The attempt at a solution

not really sure where to start. i can figure out the converging or diverging part just plug in some numbers and see if it is getting bigger (going to infinity) or if it leavels off.
not sure how to find the sum.
probaly just a simple formula but the Professor did not give it to us.

2. Mar 6, 2009

### fball558

sorry the format was lost when posted
the second problem is a division problem the -3 part on top
and the 4^n on bottom that is why the lines are there.

3. Mar 6, 2009

### yyat

Last edited by a moderator: Apr 24, 2017
4. Mar 6, 2009

### fball558

ok that would make it a lot easier
but how do you find "r" the ratio?
for example number two a(1) = 1/4
a(2) = -3/16 and a(3) = 9/64
dont know what you would do to 1/4 to get -3/16 and you have to do that same thing to -3/16 to get 9/64 right??

5. Mar 6, 2009

### yyat

You are looking for a number r such that a(1)*r=a(2), and since the series is geometric, you will also have a(2)*r=a(3) for the same r. So what is r?

6. Mar 6, 2009

### fball558

man im dumb lol i solve the first on the wrong way. got -4/3 and then -3/4 for the second that is where i messed up.
so my r would be -3/4
thanks now i can just follow the formula