Absolute Convergence of a Series

[forestmine]

Homework Statement

I'm given this series and asked whether it converges, absolutely converges, or diverges.

Ʃ(n=0 to infinity) [2(-1^n)(3^(n+1))]/5^n

Homework Equations

The Attempt at a Solution

The answer states that it is absolutely convergent, and that it converges to 15/4. Everything I've done doesn't get me to this answer. It's driving me nuts!

On the one hand, we can say it's an alternating series. So, by the alternating series test, I took the limit of b_n, and right off the bat, I can't seem to get that to equal 0, which by the alternating series test, would say that the series diverges.

So then, I tried the ratio test. I know that for the ratio test, if the limit of the ratio is < 1, the series converges. If I pull out the 2 and then use the ratio test, I do in fact get a number that is < 1, but multiplying it by 2, doesn't get me anywhere near 15/4.

I really don't know where to go from here. Obviously I'm using these tests incorrectly, but I can't see how... :/

Anyone help would be greatly appreciated!

 

[LCKurtz]

Using the ratio test to show convergence by showing the limit ratio is less than 1 is one thing. You can show absolute convergence that way. But that doesn't say anything about what the sum of the series is. Here's a hint: Factor out a 3 and think about geometric series. And show some work so we can figure out where you are going wrong.

 

[forestmine]

Ok, in that case, I'm a little confused about how to handle alternating geometric series.

I can get down to 6*(-1)^n * (3/5)^n

But the (-1) is throwing me off. I want to say that r=3/5, and a=6, and then using a/1-r, I have

6/(1-3/5) = 15.

So I suppose I'm somewhat on the right track...

 

[LCKurtz]

Ok, in that case, I'm a little confused about how to handle alternating geometric series.

I can get down to 6*(-1)^n * (3/5)^n

But the (-1) is throwing me off. I want to say that r=3/5, and a=6, and then using a/1-r, I have

6/(1-3/5) = 15.

So I suppose I'm somewhat on the right track...

Almost. Put the $-1^n$ in with the $(\frac 3 5)^n$. It's OK for the ratio to be negative.

 

[forestmine]

Ah, got it. Thank you so much!

So in general, if I'm dealing with an alternating geometric series, the (-1)^n can just be multiplied by whatever other factors I have that are raised to the n.

And as for absolutely convergent, the only test that will verify that is the ratio test, as that deals with the absolute value, correct?

What about this alternating series test? I imagine using that should also prove convergence in this case (although not a value to which the series converges). How would I take the limit of b_n. If I use L'Hopital's, I continuously wind up with a number raised to n minus some value, and thus can't seem to get the limit to equal 0.

 

[LCKurtz]

Ah, got it. Thank you so much!

So in general, if I'm dealing with an alternating geometric series, the (-1)^n can just be multiplied by whatever other factors I have that are raised to the n.

And as for absolutely convergent, the only test that will verify that is the ratio test, as that deals with the absolute value, correct?

Any test of the absolute value series that shows it converges would do. The ratio test is an obvious first thing to try.

What about this alternating series test? I imagine using that should also prove convergence in this case (although not a value to which the series converges). How would I take the limit of b_n. If I use L'Hopital's, I continuously wind up with a number raised to n minus some value, and thus can't seem to get the limit to equal 0.

Once you know it converges absolutely you can forget about the alternating series test.

For your question about $b_n$, use the fact (or prove it if you haven't seen it) that $r^n\rightarrow 0$ as $n\rightarrow \infty$ if $|r|<1$.

 

[forestmine]

Thanks again for the help.

You've definitely helped clear some things up for me. (: