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Gee. I thought they are similar.

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I can understand it better.In summary, if there is no ##(-1)^2## factor, I can find the limit. But, now I have no idea how to find limit for the ##(-1)^\infty##. I thought ##(-1)^\infty## is an indeterminate form. So, how to modify this? Thanks!f

- #36

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Gee. I thought they are similar.

- #37

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That chapter in Boas is all about tests for convergence of infinite series. I.e.infinite sums.Gee. I thought they are similar.

- #38

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Yeah, but for preliminary test, how can I proof those 1 and 2 series is divergent and series number 3 need further testing with the ##(-1)^n## and ##(-1)^{n+1}## factor? If the factor (-1)^n does not exist I can get the limit. And from this limit, I can conclude that series number 1 and 2 is divergent because the limit does not equal to 0. With the same idea, I can conclude that series number 3 need further testing because the limit equals to 0.That chapter in Boas is all about tests for convergence of infinite series. I.e.infinite sums.

So, the main problem for me is those factor.

Since I want to focus on those factor, I assume (-1)^n is a function of n. So, I ask in this thread. So sorry if my problem irrelevance with the thread's title.

- #39

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The preliminary test is whether ##a_n \rightarrow 0##. I don't know how much detail Boas goes into, but ##a_n \rightarrow 0## if and only if ##|a_n| \rightarrow 0##.Yeah, but for preliminary test, how can I proof those 1 and 2 series is divergent and series number 3 need further testing with the ##(-1)^n## and ##(-1)^{n+1}## factor? If the factor (-1)^n does not exist I can get the limit. And from this limit, I can conclude that series number 1 and 2 is divergent because the limit does not equal to 0. With the same idea, I can conclude that series number 3 need further testing because the limit equals to 0.

So, the main problem for me is those factor.

Since I want to focus on those factor, I assume (-1)^n is a function of n. So, I ask in this thread. So sorry if my problem irrelevance with the thread's title.

For the preliminary test, you can ignore the factor of ##(-1)^n##. Isn't that ironic?

In any case, it should be clear that this sequence does not converge to ##0##. Hence, the series fails the preliminary test.

To get through Boas, you are not required to prove these things using rigorous epsilon proofs. That's a different game altogether - perhaps an even harder game! Instead use the techniques that Boas provides.

- #40

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It's useful also to mention the book and the section. It makes life a lot easier if we know this information.

- #41

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The preliminary test is whether ##a_n \rightarrow 0##. I don't know how much detail Boas goes into, but ##a_n \rightarrow 0## if and only if ##|a_n| \rightarrow 0##.

For the preliminary test, you can ignore the factor of ##(-1)^n##. Isn't that ironic?

Huft. So far, I don't see that Boaz mention about the absolute value. No wonder one of PF member in another thread say that those test is for positive terms.

Okay, thanks for the suggestion how to ask in this forum.

Or should I learn back to sequence? I found a book that discuss infinite series from sequence. Or do you think learn from Boas is enough (with saving time consideration)?

- #42

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Glad that problems are already solved. Thanksss

- #43

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If you are studying mathematical methods as a physics or science major, then go with Boas. You don't want to get sidetracked into pure maths. Trust me on that.Huft. So far, I don't see that Boaz mention about the absolute value. No wonder one of PF member in another thread say that those test is for positive terms.

Okay, thanks for the suggestion how to ask in this forum.

Or should I learn back to sequence? I found a book that discuss infinite series from sequence. Or do you think learn from Boas is enough (with saving time consideration)?

Boas is an advanced book. There's everything in there up to Tensor analysis and Bessel functions. You will need to develop a lot of mathematical

- #44

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Well thanks for your reminder. I Will continue the study then . Thanks!

- #45

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Now, I will try to use the more general method: number 1.

$$\lim_{n\to \infty} (-1)^n = L$$

$$|(-1)^n - L|<\epsilon$$

$$(-1)^n <\epsilon + L$$

If ##L > 0##, then only the even number of n satisfy that condition.

If ##L < 0##, then only the odd number of n satisfy that condition.

So, the limit is DNE.

Hmm. Are my works true? Thankss

You have the correct idea and you did well for someone who isn't experienced with proofs. To improve your argument, you should state it as a proof by contradiction. As others suggest, the text you are using doesn't expect formal proofs.

You may be the sort of person who wishes to understand the details of mathematics. If so, you should study a book on Logic and understand some of the formal rules of working with the quantifiers ##\forall## and ##\exists##. I think you would find the formal study of logic easy compared to mathematical physics.

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