For the former, you can compute \(\lim_{n\rightarrow\infty}[\frac{1+(n+1)\log(n+1)}{n\log n}-1]n\) or even \(\lim_{n\rightarrow\infty}[\frac{1+(n+1)\log(n+1)}{n\log n}-1-\frac1{n^2}]n^2\). I computed the former using L'Hospital and it was 1. For the latter, let \(b_n=1+\frac1n+O(\frac1{n^2})\). Then for any \(\varepsilon>0\), and any sufficiently large \(n\) we have \(1+\frac{1-\varepsilon}n<b_n<1+\frac{1+\varepsilon}n\), which means \(e^{1-\varepsilon}<\lim_{n\rightarrow\infty}b_n<e^{1+\varepsilon}\). Now since \(\varepsilon>0\) can be arbitrarily small, the limit is equal to \(e\).
That later mess after "1+" should be epsilon. I tried to fix many times but I think the latex is not working correctly.
D'oh! Right you are, I got careless. I spent quite a bit of time since your reply looking for a sequence that worked, and I was almost ready to give up and try a different line of reasoning instead of using specific examples, but in comes Temur to the rescue Please Register or Log in to view the hidden image! For the first part, we show (as I did in my incorrect proof attempt) that if the limit superior is less than \(e\), then for large values of \(n\), we have \(\frac{1+a_{n+1}}{a_n}<1+\frac{1}{n}=\frac{n+1}{n}\). It follows then that \(a_n>\frac{n}{n+1}+a_{n+1}\left(\frac{n}{n+1}\right)\), and then we iterate this result, obtaining \(a_n>\frac{n}{n+1}+\frac{n}{n+2}+a_{n+2}\left(\frac{n}{n+2}\right)\), and so on. By using induction, we can thus show that \(a_n\geq n\left(\sum_{k=n+1}^\infty\frac{1}{k}\right)\), and since this sum is divergent, it means that \(a_n\) is not a finite term in the sequence, an obvious contradiction. Then just do what Temur did, taking \(a_n=n\mbox{ln}(n)\) to show that we can't improve our estimate of the lowest possible bound for the limit superior, and ze proof ist complete!
CptBork, I don't follow your iteration for the inequality. Nevertheless, there appears a serious problem in your inductive step. The sum in your inequality tends to 0 as n tends to infinity.
Why was my post deleted? Wasn't it scientifically correct, because... if it wasn't, why did the mod who deleted it, did so without a shift to explain his actions....?
Well what I'm arguing is that if the limit superior is less than \(e\), we can show that the sequence must contain non-finite elements, i.e. \(a_n=\infty\) for sufficiently large but finite values of \(n\). So if the limit superior is less than \(e\), we fix a value of n, sufficiently large so that \(\frac{1+a_{k+1}}{a_k}<1+\frac{1}{k}=\frac{k+1}{k}\) for all \(k\geq n\). So plugging in this inequality, we have: \(a_k>(1+a_{k+1})\left(\frac{k}{k+1}\right)=\frac{k}{k+1}+a_{k+1}\left(\frac{k}{k+1}\right)\) This inequality holds for all \(k\geq n\), so by iteration I mean that we apply it to successive values of \(k\), i.e. \(a_{k+1}>(1+a_{k+2})\left(\frac{k+1}{k+2}\right)= \frac{k+1}{k+2}+a_{k+2}\left(\frac{k+1}{k+2}\right)\) So we have an infinite string of inequalities, and we may combine them, together with the assumption that \(\{a_k\}\) is a positive sequence, to deduce that \(a_n\geq\ \sum_{k=n+1}^\infty\left( \frac{n}{k}\right)=n\left(\sum_{k=n+1}^\infty \frac{1}{k}\right)\). Hence if the assumption that the limit superior is strictly less than \(e\) were true, we would be able to find some finite value of \(n\) such that \(a_n=\infty\). Not that it makes a difference to the above argument, but I think you're incorrect about asserting that \(\lim_{n\to\infty}\sum_{k=n}^\infty\left(\frac{1}{k}\right)\) tends to zero. This sum is divergent for all finite values of \(n\), so as we let \(n\to\infty\), the limit must also be \(\infty\). Anyhow, it seems you must have used a different line of reasoning to solve the problem. In that case, I'd be delighted if you could show us how you did it.
It's not the mods' job to teach you basic mathematics. You haven't even mastered algebra which 14 year olds here in Canada are expected to know (otherwise you'd have no trouble applying the distributive law of multiplication). Take some math classes, learn how to do math, then once you've finally done that you can come here and actually contribute something of use instead of nonsensical gibberish.
No. I made mistake. The post wasn't deleted, i just muddled it in the wrong thread, just to make sure that you don't get on your high-horse here Alphanumeric.
I see the despite us having different posting styles, different posting times and different personalities, you've gone down the crank route of "Two knowledgable people disagree with me. Therefore they are the same person!". Perhaps if you posted some actual valid maths and physics we'd back off. But you don't. Because you can't.
Not quite... I have seen the time-scales in which you both have posted, and they are quite differential to about 3 minutes.
Tell you what, Reiku. If one of the moderators can verify (without actually publishing our IP addresses) that Alphanumeric and myself have two completely different IP addresses from two completely different continents, will you agree to never again post in the physics & math section, no ifs ands or buts?
And Reiku, I'm not on a high horse. You're on a high horse writing about maths that you openly admit you don't understand in the least. You're bullying your friends in the pseudoscience and parapsychology sections by trying to convince them to accept your ideas, on the grounds that you supposedly know much more than them about math & physics, when in fact you demonstrably know next to nothing.
Don't you guys have a reputation point system on this forum? It happened to me more or less when I posted my alternate hypothesis for comment and, not knowing Reiku, took his words at face value. How can something like this be avoided?
It's up to the mods what they want to do with it, so if you feel Reiku has misrepresented himself and is unfairly feeding you misinformation, you should report it to them. For my part, I call him out whenever I see him posting nonsense, it's the only thing I can personally do to try and protect newcomers and make sure they're getting accurate info. He was recently banned 3 days for making off-topic posts like what he's been doing here, if he continues to do it I'm sure it'll be more severe next time. In my humble opinion, it seems to me he's desperate to fool us that he knows anything of substance about physics and math, because he thinks if he can just luck out and convince us that he's knowledgeable, then we'll be more open to his kooky ideas about consciousness. At the same time, I'm sure it's important to tolerate Reiku's presence here as he actually hurts the case being made by those of similar beliefs. What better way to shoot a bad idea down than to show the incompetence of those who advocate it?
BTW SouthStar, I have another method of showing that our minimal estimate for the limit superior cannot be larger than \(e\). If you take \(a_n=kn\), then the limit superior is \(e^{\left(\frac{k+1}{k}\right)}\). We can make this limit as close to \(e\) as we want by choosing \(k\) to be sufficiently large, thus we can always find a sequence giving a limit superior in the range \((e,e+\epsilon)\) for any \(\epsilon>0\).
Here's an easy one: Let \(\{a_n\}\) be any strictly positive sequence such that \(\lim_{n\to\infty}a_n\) exists and is finite. Prove that \(\lim_{n\to\infty}\sqrt[n]{a_1a_2\ldots a_n}=\lim_{n\to\infty}a_n\) Then, using either this result or a certain result stated earlier in this thread by SouthStar, prove that \(\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}=\frac{1}{e}\)
Equivalent? Please Register or Log in to view the hidden image!: http://www.sciforums.com/showpost.php?p=1920220&postcount=49 The general statement is known, by the way, as the Stolz-Cesaro theorem. If \( {\lim }\limits_{n \to \infty } \frac{{x_n }}{{x_{n - 1} }}\) exists then \( {\lim }\limits_{n \to \infty } \sqrt[n]{{x_n }} = {\lim }\limits_{n \to \infty } \frac{{x_n }}{{x_{n - 1} }}\) Let \(x_n = \frac{{n!}}{{n^n }}\)
I don't consider that observation sufficient since, as temur's example shows, strict inequality does not hold.
Ah, I see now. It was my mistake. Nevertheless I think your proof is not correct. From the string of inequalities you are passing to a limit which does not exist. In analysis, we have that if a(n) and b(n) are convergent (to a and b respectively) and a(n)>=b(n) the a>=b. However, in your case, the sum b(n) is not convergent to begin with so the inequality you obtain for the series is undefined. In a phrase, the symbol \(\sum_{k=n+1}^\infty \frac{1}{k}\) is mathematically undefined. At best you've shown that the sequence a(n) must be unbounded. Recall that in mathematics, the statement that a real number is greater than infinity is undefined so if you reach that in your logic (regardless of whether it is a proof by contradiction) then you've made a mistake. That's where I made a mistake myself. Such a series would not even tend to 0 as n tends to infinity because it's not even defined. I'll let you retrace your steps and come up with a proof. If you don't get it I'll post mine.
Actually, yes now that I think of it, it is indeed equivalent to your previous statement, if we define a sequence \(\{s_n\}\) by \(s_{n+1}=s_na_{n+1}\) with \(s_1=a_1\). I never actually thought of it that way, but I guess that gives us two different ways to prove these limit theorems, since I know a different method of proving each. Also the limit theorem you stated happens to be the alternative method I was suggesting for finding \(\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}\). Now it makes sense why either approach would work for this problem, since they're really equivalent. Suppose \(\lim_{N\to\infty}\ sup\limits_{n>N}\left(\frac{1+a_{n+1}}{a_n}\right)^n\geq e+\epsilon\) for some fixed \(\epsilon>0\) and any positive sequence \(\{a_n\}\). Then by choosing a sufficiently large value of \(k\), and choosing the sequence \(a_n=kn\), I can exhibit a sequence with an even smaller limit superior, hence by contradiction, \(e\) is the best we can do for placing a lower bound on the limit superior in general. I'm trying to avoid typing the whole argument out in full detail, as it really is killer on the wrists. Regardless, we have the infinite chain of inequalities I mentioned: \(a_k>\frac{k}{k+1}+a_{k+1}\left(\frac{k}{k+1}\right)\), \(a_{k+1}>\frac{k+1}{k+2}+a_{k+2}\left(\frac{k+1}{k+2}\right)\) And so on, for \(k\geq n\), where \(n\) is a fixed integer chosen to be sufficiently large, as demonstrated in my previous post on this proof. If we combine a finite number of these inequalities, we get \(a_k>\left(\frac{k}{k+1}\right)+\left(\frac{k}{k+2}\right)+\ldots+\left(\frac{k}{k+N}\right)+a_{k+N} \left(\frac{k}{k+N}\right)\), where \(N\) is any positive integer. Then since \(\{a_n\}\) is a positive sequence, we conclude that \(a_k>\left(\frac{k}{k+1}\right)+\left(\frac{k}{k+2}\right)+\ldots+\left(\frac{k}{k+N}\right)\), for arbitrary positive integers \(N\). Now set \(k=n\), where \(n\) is some sufficiently large, fixed finite integer whose existence was demonstrated earlier, and let \(N\to\infty\). The conclusion then follows directly. As \(N\to\infty\) in the line above, we conclude that \(a_n\) is larger than any finite number, even though \(n\) is itself a fixed finite integer.
