Is this the right way to analyze whether the series converges or not?

1. Apr 9, 2013

Lo.Lee.Ta.

1. series: 3 - 1 + (1/3) - (1/9) + (1/27) - ...

2. Since it looks like the denominator just gets multiplied by 3 each time (excluding the first 3),
would I say that the equation is:

3 - (1/(3n)) ???

And then I would say that the (1/(3n)) is my r?

1/(1 - (1/3)) = 3/2

So that means it converges to 3/2? Is that right...? :/

Thanks so much for the help! :D

2. Apr 9, 2013

Curious3141

There are many tests for series convergence. However, you don't need them here, because you should be able to recognise that this is a geometric series, with first term $3$ and common ratio $r = -\frac{1}{3}$. Every geometric series with $|r|<1$ converges. This is an immediate consequence of the ratio test for convergence.

Can you see this? Divide each term by the one before and verify that the ratio is constant.

Do you know the formula for the sum of a geometric series?

3. Apr 9, 2013

Lo.Lee.Ta.

Okay, so I need to include the - and say that r= -1/3.

Is my method incorrect for finding the r, though?

1/(1 - (-1/3)) = 3/4

So does it converge to 3/4?

Yeah, I'll also have to remember that if |r|<1, it converges.

4. Apr 9, 2013

Curious3141

Don't forget the first term is NOT 1. It's 3.

So the sum should be $\displaystyle \frac{3}{1 - (-\frac{1}{3})} = \frac{9}{4}$

5. Apr 9, 2013

Lo.Lee.Ta.

Oh, right! a is the first term- and that's 3!

Thanks so much for telling me! :)

6. Apr 10, 2013

Lo.Lee.Ta.

Okay, now I'm wondering- whenever you have an alternating positive and negative series, will the r always be negative?

In our situation, the equation was: 3 - (1/3n)

Will it always be minus there when it's an alternating + and - series?

Will it ever be: 3 + (1/3n)?

Thanks! :)

7. Apr 10, 2013

Staff: Mentor

r represents the common ratio in a geometric series. If the terms in this series alternate in sign, then r necessarily is negative.