Alternating Series, Testing for Convergence

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• Drakkith
In summary: Thus, the whole thing converges.In summary, the criteria for testing for convergence with the alternating series test is that the series must have the form Σ(-1)n-1bn, with bn>0, bn+1 ≤ bn for all n, and lim n→∞bn = 0. However, it has been noted that the sequence may not always strictly follow these criteria, but as long as the sequence is eventually decreasing and converges to zero, the series will still converge. This has been confirmed by comparing the behavior of the first N terms (where N is any fixed finite number) and observing that the series is alternating and eventually decreasing, allowing for the application of the alternating series test.

Drakkith

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

My question is about the criteria. I'm running into several homework problem where bn is not always greater than bn+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 bn < bn+1. But my criteria says bn+1 ≤ bn for all n.

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

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.

Drakkith
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
axmls said:
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.

axmls said:
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.

pwsnafu said:
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.

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.

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Drakkith

1. What is an alternating series?

An alternating series is a series in which the terms alternate in sign between positive and negative. For example, 1 - 1/2 + 1/3 - 1/4 + ... is an alternating series.

2. How do you test for convergence of an alternating series?

The most commonly used test for convergence of an alternating series is the Alternating Series Test. This test states that if the terms of the series decrease in absolute value and approach 0, the series will converge. Additionally, the series must also satisfy the condition that the absolute value of the terms is eventually less than or equal to the absolute value of the first term.

3. Can an alternating series converge to a non-zero value?

Yes, an alternating series can converge to a non-zero value. This happens when the absolute values of the terms do not approach 0, but rather approach a non-zero value. In this case, the series will converge to the sum of the alternating terms.

4. How can you determine the sum of an alternating series?

The sum of an alternating series can be determined by using the formula S = a - b, where S is the sum, a is the sum of the positive terms, and b is the sum of the negative terms. This formula is derived from the fact that the terms of an alternating series can be grouped into pairs, where the first term is positive and the second term is negative, and the sum of each pair is equal to a - b.

5. Are there any other tests for convergence of alternating series?

Yes, there are other tests for convergence of alternating series such as the Ratio Test, Root Test, and Integral Test. However, these tests are not specific to alternating series and can be applied to any series. The Alternating Series Test is specifically designed for alternating series and is often the most efficient method for testing convergence in these types of series.