Solving a Limit to Determine the Convergence of a Series

[BraedenP]

Homework Statement

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}

Homework 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).

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...

 

[Mark44]

Homework Statement

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}

Homework 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).

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...

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}

 

[BraedenP]

Oh, okay. That makes sense. Thanks!

 

[SammyS]

Okay.. That makes sense, but in this case the limit evaluates to infinity, doesn't it? It should evaluate to 0.

\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.

 

[Mark44]

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

 

[BraedenP]

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