Is there actually a difference if I write the series (-1)^(n)/(n-n^(1/2)) as (-1)^(n+1)/(n-n^(1/2))? The only difference is the coefficient on -1, instead of n being n+1... it shouldn't make a difference right? Wolframalpha gives me different results.
@Clash: this changes your sum by more than 2/(2 - sqrt(2)) because the first term in your sum (for n=2) is 1/(2-sqrt(2)) in one case and -1/(2-sqrt(2)) in the other case. What do you think, @Srivatsan?
@HenningMakholm I believe it might be a language problem. Perhaps he wanted to say: I need also the proof, not only the explanation. And he does not speak English good enough.
We have here many posts that have language problems.
@AsafKaragila The only thing I was able to find was this one: math.stackexchange.com/questions/38900/binomial-expansion It's not a duplicate, since it asks about a specific method of proving this identity. Either I should try to improve my searching skills, or it is really not on math.SE.
@Clash: Martin's answer is perhaps the clearest and easiest to write as a proof; Fezvez's answer matches my intuition more clearly. In the long run, I think Fezvez's approach is easier to see what's going on. I suggest you understand the approach if not precise estimates used.
[For completeness sake, let me add that I don't understand Davide's answer is coming from. =)]
when calculating this limit: lim n->infinity (-1)^(n+3)*(n+2)/(5n^2+1)^(1/2), can I remove the alternating part? and say that it goes to result * -1 or result * 1? For that case, 1/sqrt(5), so 1/sqrt(5) or -1/sqrt(5)
therefore the series diverges, as lim n -> infinity != 0?
I commented on someones answer "that's wrong because .." then he said something and I said "oh my mistake I thought you were talking about X but you really meant Y"
now they trying hard to find mistakes in my answer ...
First note that $$\lim_{x \to \tfrac{\pi}{2}^-} \sin(x) = 1$$ and $$\lim_{x \to \tfrac{\pi}{2}^-} \cos(x) = 0.$$
therefore $$\lim_{x \to \tfrac{\pi}{2}^-} \frac{\sin(x)}{\cos(x)} = + \infty$$ since we have (by the definition of limit):
$$\forall \varepsilon_1, \exists \delta_1, \forall x < \...
so I got distracted with all that notational rubbish, when all he was really saying is that we should be considering one sided limits (which is correct)
I was confused because it looked like you commented on your own answer saying that you mistook yourself and as a result, other people tried to disprove your answer
The way I wrote it was: A and B therefore C -- <definition of A> and <definition of B> therefore <definition of C> by [proof method]: verification: showing that proof method actually works here
guys is there any sequence a_n, where lim n -> infinity a_n = 0 and \sum (-1)^n*a_n diverges? I wanted to use something like a_n=2^(1/n), but that clearly isn't going to get me lim = 0...
In mathematics, the root test is a criterion for the convergence (a convergence test) of an infinite series. It depends on the quantity
:\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|},
where a_n are the terms of the series, and states that the series converges absolutely if this quantity is less than unity but diverges if it is greater than unity. It is particularly useful in connection with power series.
The test
The root test was developed first by Augustin-Louis Cauchy and so is sometimes known as the Cauchy root test or Cauchy's radical test. For a series
:\sum_{n=1}^\infty a_n.
the...
@Srivatsan but I haven't done anything wrong this time, right? It's because on the previous exercises we could only use the comparison and the alternate series test. and that's more stuff to write
@QED Maybe, but you gave him a chance to. And it's not like anything is lost, right? You can rectify the mistake, improve the answer? At the least you can ignore the comments =)
for the alternate series test, we have to show lim = 0, monotone falling and a_k > a_k+1... for the comparison test, well you have to compare it, that's more stuff to write
@mercio what do you mean with basic logic? I just mean we couldn't use for example the ratio test, because we hadn't learned it... if by logic you could reach that test, you still wouldn't be allowed... and yes I also disagree with this, but well, im not the professor
@Srivatsan oh ok, well, we know that k! < k^k. we know that the root test (k-th root) for k^k is k, and that the root test for k! is smaller than k. then we have something smaller than k divided by k => it's smaller than 1 => it converges by the root test
@Clash That ain't enough. What does the root test give for the harmonic series? For the series sum 1/n^2? The test is that the nth root of a_n should be bounded away from 1. That is, it should be at most some k (independent of n) that is strictly smaller than 1.
@QED yeah this is exactly what I mean, 5!^1/5 < 5, and then we have something smaller than 5 / 5, which is smaller than 1. by the root test it converges then
@Clash no I'm afraid the root test is "if nth root of un < a < 1 for all n, then the series converge" and not "if nth root of un < 1 for all n, then the series converge", which is false because the harmonic series diverges for example
@Clash Incorrect. If a_n = n!/n^n, then it is true that a_n^{1/n} < 1, but that is not enough. (*Think* more carefully about the harmonic series example.)
@Clash Yes, that shows that n!/(n^n) is less than 1. But as you take the limit, you can only conclude that lim n!/ (n^n) [if it exists at all =)] is less than or equal to 1.
@Srivatsan yeah, my proof is not about saying that if each element is < 1, than it converges. As 1/n diverges and each term is smaller than 1. But the test says, if you take the n-th root and make the limit n->infinity, if it's smaller than 1, it converges. have I done something wrong? really sorry if im repeating myself and didnt understand u guys
ohhhhhh equal to 1
but why equal to 1? I mean k! is always (k>1) smaller than k^k
That's a good question. Think about the two simpler examples where it's clear what's going on. (1) 1 - 1/n. (2) 1/n^{1/n}.
If the sequence can get arbitrarily close to 1 (and the limit exists), then the limit has to be 1. No other go. =)
That's the case with 1/n^{1/n} and 1-1/n. True that they are small than 1 individually, but the sequence--as a unit--gets close to 1, and that's all that matters for the limit to be equal to 1.
That doesn't happen with the sequence n!/n^n, but that's not a trivial statement. Needs some proof. [If you are interested, the limit of n!^{1/n} / n is equal to 1/e, which is strictly smaller than 1 and you are done. But establishing this formula is a bit involved.]
My god, it's like Bill was reading the comments in the wrong order but he then insists that he doesn't care. I guess I should have guessed that the OP just wanted to see a proof and understand it on his own... before even saying so.
I guess asking on chat is the easiest way: Can I find 2 more people voting here for reopen - math.stackexchange.com/questions/86093 (And then 5 votes for closing the question will be needed - but this time as exact duplicate, not too localized.)
@AsafKaragila I guess after reopening and closing as an exact duplicate, many comments from that discussion will be obsolete, so we can perhaps delete some of them.
Even if only out of respect to Bill Dubuque - if he likes this solution better - reopening and closing is worth the trouble.
@QED If I may quote Michael's comment: _ I've voted to re-open. In the notice explaining that it was closed, it says "This question is unlikely to ever help any future visitors". That doesn't make any sense at all. It seems as if closure resulted from people misunderstanding the poster's comments as meaning he doesn't want to understand the proof but just wants to copy it so he can turn it in. _
- If that is not the interpretation, what is? I can't make anything out of a person coming and asking me to prove a random identity. It's (1) homework, (2.) he is interested in understanding in which case the OP's comment makes no sense, (3.) he wants to get his own understanding from a formal proof.
"For example in 𝔸^2_ℚ, the maximal ideal M=⟨x^2−2,y⟩⊂k[x,y] yields a closed point which is not rational (according to a theorem proved 25 centuries ago, albeit not in the language of scheme theory). "
@robjohn Unimaginative solution. 1. Get the permalink, 2. Copy the message number, the number following the #, 3. Type : and then paste the number, followed by your actual comment.