The limit of the function as n approaches infinity, specifically the expression ##\lim_{n\to\infty} \frac{(-1)^n n^2}{(n+1)^2}##, does not exist due to the oscillatory nature of the sequence ##(-1)^n##. The discussion emphasizes that terms like ##(-1)^\infty## are fundamentally incorrect and should be avoided. Instead, it is established that the sequence oscillates between -1 and 1, leading to the conclusion that the limit does not converge to a finite value. Proper mathematical proof techniques, including the epsilon-delta definition of limits, are necessary to substantiate these claims.
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Boas does not cover things like the non-convergence of the sequence ##(-1)^n##, or rigorous real analysis. I suspect you are just supposed to assume that ##(-1)^n## "obviously" does not converge to anything.agnimusayoti said:I study Mathematical Methods for Physical Sciences and uses ML Boas book. Nope, I get the notation n and N in the youtube, but the presenter do not explain what is the difference between those two number and their relevance to the proof. Apologize for my lack of knowledge.
These are infinite series, not sequences! That's a completely different thing.agnimusayoti said:From Boas, in Problem section 5. There are 3 problem that I failed to find the limit of the sequence because of ##(-1)^{n+1}## and ##(-1)^n## factor. I think if I can find the limit of ##(-1)^n## as n approaching infinity there will become a good experience for my learning process.
For the detail, the sequence are:
1. ##a_n=\frac{(-1)^{n+1} n^2}{n^2 + 1}##
2. ##a_n=\frac{(-1)^{n} n^2}{(n + 1)^2}##
3. ##a_n=\frac{(-1)^n n}{\sqrt{n^3 + 1}}##
But, interestingly, if I neglect the ##(-1)^n## factor, I can use the preliminary test and therefore solve the problem.
That chapter in Boas is all about tests for convergence of infinite series. I.e.infinite sums.agnimusayoti said:Gee. I thought they are similar.
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.PeroK said:That chapter in Boas is all about tests for convergence of infinite series. I.e.infinite sums.
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##.agnimusayoti said: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.
PeroK said: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?
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.agnimusayoti said: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)?
agnimusayoti said: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