# Homework Help: Solving a Limit to Determine the Convergence of a Series

1. Sep 16, 2011

### BraedenP

1. The problem statement, all variables and given/known data

I am asked to determine whether a series converges, and if so, to provide its sum.

The problem is:

$$\sum_{n=1}^{\infty}(-3)^{n-1}4^{-n}$$

2. Relevant equations

- I know that if the limit of the sequence as n->inf is finite, then the series converges at that limit.

- I also know that the sum of the given series is equal to said limit (if it exists).

3. The attempt at a solution

1. I rewrote the series as a limit and turned it into a fraction: $$\lim_{n\rightarrow\infty}\frac{(-3)^{n-1}}{4^n}$$

2. Then as $n\rightarrow\infty$ the denominator approaches infinity, and the numerator begins to oscillate (for even n values, it's a large positive number, and for odd n values it's a large negative number).

3. Thus, I concluded that the limit does not exist, and thus the series does not converge.

However, this limit actually evaluates to 0, meaning that the series converges and sums up to 0. How is this? I'm quite confused...

2. Sep 16, 2011

### Staff: Mentor

Pull a factor of 1/4 out to get the exponents aligned, so that what you're summing is $$\left(\frac{-3}{4}\right)^{n-1}$$

3. Sep 16, 2011

### BraedenP

Oh, okay. That makes sense. Thanks!

Last edited: Sep 16, 2011
4. Sep 16, 2011

### SammyS

Staff Emeritus
$\displaystyle \lim_{n\to\infty}\left(\frac{-3}{4}\right)^{n-1}=0$

This is necessary for the series (the sum) to converge, but it's not enough to guarantee it.

That will take a bit more work.

As a start, write out the first several terms ot the series, then look at the sum of any two consecutive terms of the series.

5. Sep 16, 2011

### Staff: Mentor

Alternatively, this is a geometric series, which is probably one of the first series presented when you started learning about series.

6. Sep 16, 2011

### BraedenP

Yeah, thanks guys. I was reading that all wrong. Makes perfect sense now :)