Geometric Series: 2/3^k and -2/10^k

[Kyousik]

Homework Statement

(infinity)sigma(k = 0) [2(2/6)^k + (-2/10)^k)

Homework Equations

Geometric Series

The Attempt at a Solution

I split these up into two geometric series

(infinity)sigma(k = 0) [2(1/3)^k]
2 / (1 - 1/3)
r = 3

This diverges.

(infinity)sigma(k = 0) (-1/5)^k
1 / (1 + 1/5)
r = 5/6

This converges.

Now, I'm not sure what to do? If it does converge, I need to find its sum .

 

[StatusX]

The first series doesn't diverge. What makes you think it does?

 

[Kyousik]

Oh oops! I see, the |r| value is the ratio. Not the number you get after. The number you get after is what the series converges to?

So 3 + 5/6 = 18 / 6 + 5 /6 --> 23 / 6

Would that be correct?

 

[StatusX]

Yea. Although to be more rigorous, you need to show those two series converge absolutely to justify splitting the sum. This is only a problem for the second one, which is an alternating series, but taking the absolute value of each series gives another converging series, so you're ok.

This might sound like a technicality, and it is, but there are examples where your method would give a wrong answer. For example, it wouldn't work on the series:

\sum_{n=1}^\infty ((-1)^n+(-1)^{n+1})

There are also examples where it would give you a finite answer, but which would be wrong. These are rare, and I can't think of one off hand, but I'll post one if I do.

 

[Kyousik]

So to prove that, I would use the ratio test to show absolute convergence? Absolute convergence was very confusing reading from the textbook.

I forgot there were other ways to show absolute convergence. Like using harmonic, or comparison or something.

 

[StatusX]

A series \sum a_n converges absolutely iff \sum |a_n| converges. If the series is already positive, like the first one, you're done. Since the second isn't, you need to show that by taking the absolute value, which just amounts to changing -1/5 to 1/5, you still get a series that converges.