Alternating Series, Testing for Convergence

[Drakkith]

The criteria for testing for convergence with the alternating series test, according to my book, is:
Σ(-1)^n-1b_n
With b_n>0, b_n+1 ≤ b_n for all n, and lim n→∞b_n = 0.

My question is about the criteria. I'm running into several homework problem where b_n is not always greater than b_n+1, such as the following: Σ(-1)ⁿ sin(6π/n). This sequence is also not always greater than zero either (n=4 and n=5 make this negative), nor is it (-1)^n-1 like the criteria says, but the series converges anyways.

From n=6 to n=12, it appears that b_n < b_n+1. But my criteria says b_n+1 ≤ b_n for all n.
Should this read as n→∞ instead of for all n?

Am I missing something? What's with these apparent inconsistencies?

 

[axmls]

That's odd. My textbook also says for all $n$, however I checked Paul's notes online and he specifically points out that it only needs to be eventually decreasing, and your sequence does eventually strictly decrease.

I would go with Paul's notes. After all, suppose that the sequence $a_n$ is not decreasing for $1, 2, ... N$ but that it is decreasing for all $n > N$ (and further suppose that $\lim_{n \to \infty} a_n = 0$). Then certainly we could simply write the sum as $$\sum _{n=1} ^\infty (-1)^n a_n = \sum_{n=1} ^N (-1)^n a_n + \sum_{n = N+1} ^\infty (-1)^n a_n$$ Then certainly the first term is finite, and the second term converges by the alternating series test.

Of course, you'd have to show that your sequence is in fact strictly decreasing after some $N$, but intuitively that's certainly the case for $\sin(x)$ as $x \to 0$. In this case, I'd say your function is strictly decreasing for $n \geq 12$. I'd love to hear someone else's opinion, though.

It's quite possible that the textbook intends to say this: get the series in a form such that it is always decreasing, even if you have to split it up into some finite sum and an infinite sum.

 

[pwsnafu]

If you can show $\sum_{n=13}^\infty (-1)^n b_n$ converges, then trivially $\sum_{n=1}^\infty (-1)^n b_n$ also converges because all you are doing is adding a finite number of terms. Also $(-1)^{n} = - (-1)^{n-1}$ so it's the same thing.

 

[Drakkith]

My textbook also says for all n, however I checked Paul's notes online and he specifically points out that it only needs to be eventually decreasing, and your sequence does eventually strictly decrease.

That's what I figured. It wouldn't make sense otherwise.

It's quite possible that the textbook intends to say this: get the series in a form such that it is always decreasing, even if you have to split it up into some finite sum and an infinite sum.

It's possible. I'll read it over again and see if I missed something.

If you can show $\sum_{n=13}^\infty (-1)^n b_n$ converges, then trivially $\sum_{n=1}^\infty (-1)^n b_n$ also converges because all you are doing is adding a finite number of terms. Also $(-1)^{n} = - (-1)^{n-1}$ so it's the same thing.

Okay. That was my train of thought, but I didn't know if there was some crucial difference I was missing. Thanks.

 

[jbunniii]

The behavior of the first $N$ terms (where $N$ is any fixed finite number) has no effect on whether the series converges or diverges. (Of course, those terms do affect the value to which the series converges, if it converges.) Observe that $\sin$ is nonnegative and monotonically increasing on $[0,\pi/2]$. For $n \geq 12$, we have $6\pi/n \in [0,\pi/2]$. This means that we are in the domain where $\sin$ is monotonically increasing, so $6\pi/(n+1) < 6\pi/n$ implies $0 < \sin(6\pi/(n+1)) < \sin(6\pi/n)$ for $n \geq 12$. Also,
$$\lim_{n \to \infty}\sin(6\pi/n) = \sin\left(\lim_{n \to \infty} 6\pi/n\right) = \sin(0) = 0$$
where we can bring the limit inside $\sin$ because $\sin$ is continuous.

This means that for $n \geq 12$, the series is alternating and so the alternating series test applies.

Putting it slightly more formally:

$$\begin{aligned}
\sum_{n=1}^{\infty} (-1)^n \sin(6\pi/n) &= \sum_{n=1}^{11}(-1)^n \sin(6\pi/n) + \sum_{n=12}^{\infty}(-1)^n\sin(6\pi/n) \\
&= \sum_{n=1}^{11}(-1)^n \sin(6\pi/n) - \sum_{m=1}^{\infty}(-1)^{m}\sin(6\pi/(m+11)) \\
\end{aligned}$$
In the last expression, the first sum is taken over finitely many terms, so of course it converges. The second sum is an alternating series which converges as discussed above.