# Geometric Series

1. Mar 10, 2010

### mattmannmf

{sigma} 2*(-1/4)^(n-1)

Could i treat this as a geometric series? i know geometric is in the form of ar^n but the n is (n-1)

my A=2
my r= -1/4

2. Mar 10, 2010

### Staff: Mentor

$$\sum_{n = 1}^{\infty} 2 (-1/4)^{n - 1}$$

Write a few terms of this series and see if you can adjust the starting index.

3. Mar 10, 2010

### mattmannmf

i have no idea what you mean by that

4. Mar 10, 2010

### Hurkyl

Staff Emeritus
You have
2(-1/4)n-1
and you want
A rn-1
?

Why not set them equal and try solving for A and r?

5. Mar 10, 2010

### mattmannmf

no... im trying to solve whether or not the series converges or diverges and then i have to determine the sum of the series... meaning where the series sums up to..

6. Mar 10, 2010

### Hurkyl

Staff Emeritus
Oh, my mistake. I thought the method you had wanted to use was to rewrite the summand in a standard form, and were stuck on how to do that rewrite.

7. Mar 10, 2010

### Char. Limit

Well, just change n-1 into n using exponent rules.

8. Mar 10, 2010

### Staff: Mentor

Write a few terms of the series and you should quickly see that it is a geometric series. A geometric series is one for which each term is a constant multiple of the previous term.

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