# B Absolute convergence Text Book question: Boas 3rd Ed

1. Oct 22, 2016

### Battlemage!

In Mary L. Boas' Mathematical Methods in the Physical Science, 3rd ed, on page 17 it goes over absolute convergence, and defines the test for alternating series as follows:

An alternating series converges if the absolute value of the terms decreases steadily to zero, that is, if |an+1| ≤ |an|, and limn→∞ an = 0.​

Now, my confusion starts with the limit. In the text book, the limit is NOT of the absolute value of an, otherwise they would have written |an|, right?

In the example given to show this definition in operation, they are using the series ∑ [(-1)n+1]/n from 1 to infinity, however the limit they compute is limn→∞ 1/n and NOT limn→∞ [(-1)n+1]/n.

So did they just forget the absolute value sign in the definition (where they wrote the limit)? Or did they just feel showing the absolute value was trivial? Or do I just not understand the definition given? Does an always mean you just disregard the (-1)n?

Thanks for insight.

2. Oct 22, 2016

### pasmith

$\lim_{n \to \infty} a_n = 0$ if and only if $\lim_{n \to \infty} |a_n| = 0$ so in that sense it doesn't matter.

3. Oct 22, 2016

### Battlemage!

Thought so. While testing power series end points I've been including the (-1)n terms in doing the limit test to see if they diverge, but this will save me some time. For example, I had one series which was (-1)nn3x3, and when testing the boundary point x = -1, I could have just looked at n3 instead of worrying about -1. Granted, it wasn't really much of a time saver since it's obvious that (-1)n*(-1)n is always positive, but I can see where it might help.

I appreciate the info.

4. Oct 22, 2016

### Battlemage!

I do have one other somewhat related question. For the integral test, can I also do the same with the (-1)n term? That would make things easier. For example, say I wanted to use the integral test to see if (-1)n/n2 converged. Could I just integrate 1/n2?

I mean since an infinite integral is really a limit, right?

EDIT- I'm thinking NO, since 1/n diverges but (-1)n+1/n converges.

5. Oct 22, 2016

### Staff: Mentor

Besides that, you would have to explain what $\int_0^{\infty} \frac{(-1)^{x+1}}{x}dx$ actually means.
Yes, $\int_0^{\infty} f(x) dx = \lim_{a \rightarrow \infty} \int_0^{a} f(x) dx$