Proper form for geometric series with a neg inside

[Jbreezy]

Homework Statement

Curious about this ...I have to find the sum.

Homework Equations

The Attempt at a Solution

Ʃ (1/4)(-1/3)^n from 1 to infinity

I want to know the proper form and why. Is it (1/4) Ʃ (-1/3)(1/3)^n-1 or (1/4) Ʃ (1/3)(-1/3)^(n-1)

You get different answers

 

[Mark44]

Homework Statement

Curious about this ...I have to find the sum.

Homework Equations

The Attempt at a Solution

Ʃ (1/4)(-1/3)^n from 1 to infinity

I want to know the proper form and why. Is it (1/4) Ʃ (-1/3)(1/3)^n-1 or (1/4) Ʃ (1/3)(-1/3)^(n-1)

You get different answers

What's the point of bringing out a factor of 1/3? You did the same thing in your other thread.

In the above, your first formula is incorrect. (-1/3)(1/3)^n-1 = (-1/3)[1 + 1/3 + 1/9 + ...

For a geometric series such as $\sum_{n = 1}^{\infty}r^n$, if |r| < 1, the series converges to r/(1 - r).

Here's a simplistic explanation:
Let S = r + r² + ... + rⁿ + ...
Then -rS = -r² - r³ - ... - r^{n + 1} - ...

Then S - rS = r, or S(1 - r) = r, so S = r/(1 - r)

 

[Jbreezy]

I'm not sure we take out the factor in class because that is how it is written in general in the book formula.

I thought the sum of a geometric series was S = A / (1 -r) where A is the coefficient in front r. Not r/ (1-r)

 

[Ray Vickson]

I'm not sure we take out the factor in class because that is how it is written in general in the book formula.

I thought the sum of a geometric series was S = A / (1 -r) where A is the coefficient in front r. Not r/ (1-r)

Go back and read it again. The exact formula to use depends on whether you have the sum $\sum_{n=0}^{\infty} A r^n$ or $\sum_{n=1}^{\infty} A r^n$. Do you notice the difference?

 

[Jbreezy]

Yeah n = 0 or n = 1. Your first one is OK your second one isn't it needs to be n-1