Series Convergence/Divergence Problem

[AxeluteZero]

Homework Statement

Does the series converge or diverge? Give reason for your answer; if it converges, find its sum.

$$\infty\sum$$ n=0 $$\frac{(-1)^{(n+1)} * 3 - 1}{2^n}$$

Homework Equations

If |r|<1, the geometric series converges to a/(1-r). If |r|> or = 1, it diverges.

The Attempt at a Solution

$$\infty\sum$$ n=1 $$\frac{(-1)^{(n+1)} * 3}{2^n}$$ <---- This is a very similar problem that I was able to figure out.

In this problem, it's a geometric series that converges to 1 (with a sum of $$\frac{(3/2)}{1-(-1/2)}$$.

However, this particular problem has two differences. One is the -1 on top, which shouldn't matter as n-> infinity since it's so small. The other difference is it starts at zero, so I'm not sure if the above equation (in part "b") is relevant, with a/(1-r). I tried finding an "a" and got -4, but I'm not sure what to do about the r, besides assuming that the 3 and -1 don't matter:

Then I'd get (-1)^n+1 / 2^n which would be 1/2 = r. Help?

 

[Mark44]

Fixed your LaTeX. You can see what I did by double-clicking the summations.

Homework Statement

Does the series converge or diverge? Give reason for your answer; if it converges, find its sum.

$$\sum_{ n=0}^{\infty}\frac{(-1)^{n+1} * 3 - 1}{2^n}$$

Homework Equations

If |r|<1, the geometric series converges to a/(1-r). If |r|> or = 1, it diverges.

The Attempt at a Solution

$$\sum_{n = 1}^{\infty}\frac{(-1)^{n+1} * 3}{2^n}$$ <---- This is a very similar problem that I was able to figure out.

In this problem, it's a geometric series that converges to 1 (with a sum of $$\frac{(3/2)}{1-(-1/2)}$$.

However, this particular problem has two differences. One is the -1 on top, which shouldn't matter as n-> infinity since it's so small. The other difference is it starts at zero, so I'm not sure if the above equation (in part "b") is relevant, with a/(1-r). I tried finding an "a" and got -4, but I'm not sure what to do about the r, besides assuming that the 3 and -1 don't matter:

Then I'd get (-1)^n+1 / 2^n which would be 1/2 = r. Help?

 

[AxeluteZero]

Ah, thank you!

 

[Mark44]

I would expand the summation to see how close it is to the summation you've already worked with.

 

[AxeluteZero]

In terms of the series itself, the ACTUAL series I'm working with (with the extra -1) gets closer to zero more slowly than the other.

I can't use the comparison, ratio/root, or integral tests on this either. Only the geometric series formula.

So, in other words...

The top term oscillates between 2 and -4 and the bottom goes to infinity, thus Lim (n -> infinity) = 0 so the series converges to SOME number, correct so far?

 

[Mark44]

If you expand the series you'll see that it is a geometric series. What do the first four or five terms of this series look like?