Find the limit \(\frac{1}{{\sqrt n }}|\sum\limits_{k = 1}^n {( - 1)^k \sqrt k } |\) - For high school students
Is the OP overshooting his audience, or are there common high school courses which would allow one to rigorously derive for large n, \(\frac{1}{\sqrt{n}}\sum_{k=0}^n \sqrt{k} \approx \frac{2}{3} n + \frac{1}{2} + O(\frac{1}{\sqrt{n}}) \) ? That step is necessary for one straightforward way to solve this problem. Or you can jump to derivatives of integrated terms or integrations of derivatives, but how would a high school student be expected to know such results have rigor unless they have mastered the subject, which is not in my experience a high school topic.
I'm thinking this can be turned into a Riemann integration problem, or at least compared to one. My plan is to try splitting the series into two parts and summing the even and odd terms separately after multiplying by an additional factor of 1/n, so now we're working with the difference of two different, convergent series. Not sure yet if I'll be able to do this, but I think so. If I can express the resulting two series as integrals and get a finite difference, then when I don't multiply by 1/n before taking the limit, which returns us to the original problem, I would get an answer of "\(\infty\)".
Taylor series should be elementary. \(\sqrt{k+\delta} = \sqrt{k}+\frac{\delta}{2\sqrt{k}} -\frac{\delta^2}{8 k\sqrt{k}}+ \dots \) Change of variables of sum, substitution of the Taylor series, followed by integration and evaluating the limit: \(\lim_{n\to\infty} \left | \sum_{k=0}^n \frac{(-1)^k \sqrt{k}}{\sqrt{n}} \right| = \dots = \lim_{n\to\infty} \frac{1}{\sqrt{n}} \sum_{k=0}^n \frac{1}{4\sqrt{k}} + \frac{1}{32 k^{\frac{3}{2}}} + \dots = \lim_{n\to\infty} \frac{1}{\sqrt{n}} \int_{0}^n \frac{1}{4}x^{-\frac{1}{2}} + \frac{1}{32} x^{-\frac{3}{2}} + \dots dx = \lim_{n\to\infty} \frac{\frac{1}{2}\sqrt{n} - \frac{1}{16\sqrt{n}} + \dots}{\sqrt{n}} = \frac{1}{2}\) Or you can write: SUM = EVENS - ODDS EVENS = HALF ODDS = WHOLE - EVENS = WHOLE - HALF then you have SUM = EVENS - ODDS =2 HALF - WHOLE or \(\lim_{n\to\infty} \left | \sum_{k=0}^n \frac{(-1)^k \sqrt{k}}{\sqrt{n}} \right| = \lim_{n\to\infty} 2 S(n/2) - S(n)\) Where \(S(n) = \sum_{k=0}^n \sqrt{\frac{k}{n}} = \frac{2}{3} n + \frac{1}{2} + ...\) And so \(\lim_{n\to\infty} \left | \sum_{k=0}^n \frac{(-1)^k \sqrt{k}}{\sqrt{n}} \right| = \lim_{n\to\infty} 2 \left[ \frac{2}{3} \frac{n}{2} + \frac{1}{2} + ... \right] - \left[ \frac{2}{3} n + \frac{1}{2} + ... \right] = \frac{1}{2} \)
Hi rpenner, I realized after a bit of simple analysis that it's impossible to get a nonzero value for the integral by my method (multiplying by 1/n and hoping the anser is still finite), so if the limit had indeed been infinite, I wouldn't have been able to prove it using this method anyhow. I decided to plug it into Matlab and indeed it looks like the limit is 1/2. So I fooled around a little more, and it turns out that after a small bit of manipulation, I can indeed use Riemann summation to convert the problem as is into an integral. That would make it a high-school level problem in terms of required background, and when I was in high school I had seen a few problems like this during independent study, but this kind of stuff was way too advanced for our calculus class, and went far beyond our mediocre government's educational ambitions. I did it in my head and got 1/2 as an answer (knowing of course that's what I wanted to get), but I'll type out a rigorous argument so everyone can check it.
Ok, here's my proof: \(\lim_{n\to\infty}\frac{1}{\sqrt{n}}\sum_{k=1}^n (-1)^k\sqrt{k}=\lim_{n\to\infty}\frac{1}{\sqrt{2n}} \sum_{k=1}^{2n} (-1)^k\sqrt{k}=\lim_{n\to\infty}\frac{1}{\sqrt{2n}} \sum_{k=1}^n\left(\sqrt{2k}-\sqrt{2k-1}\right)=\lim_{n\to\infty}\left(\frac{\sqrt{n}}{n\sqrt{2}}\right) \sum_{k=1}^n \frac{1}{\left(\sqrt{2k}+\sqrt{2k-1}\right)}\) Note that the last equality is derived by manipulating the \(\frac{1}{\sqrt{n}}\) factor and by rationalizing the numerator. Now we see that in this form, the limit can be put into something closely resembling a Riemann sum. With a bit of algebra we get: \(\lim_{n\to\infty}\frac{1}{\sqrt{n}}\sum_{k=1}^n (-1)^k\sqrt{k}=\frac{1}{2}\ \lim_{n\to\infty}\frac{1}{n} \sum_{k=1}^n \frac{1}{\left(\sqrt{\frac{k}{n}}+\sqrt{\frac{k}{n}-\frac{1}{2n}}\right)}\) Now this would be a Riemann sum if only we could somehow get rid of that pesky \(\frac{1}{2n}\) term. We can easily demonstrate the following chain of inequalities: \(\frac{1}{2}\ \lim_{n\to\infty}\frac{1}{n} \sum_{k=1}^n \frac{1}{\left(\sqrt{\frac{k}{n}}+\sqrt{\frac{k}{n}}\right)}\leq \frac{1}{2}\ \lim_{n\to\infty}\frac{1}{n} \sum_{k=1}^n \frac{1}{\left(\sqrt{\frac{k}{n}}+\sqrt{\frac{k}{n}-\frac{1}{2n}}\right)} \leq \frac{1}{2}\ \lim_{n\to\infty}\frac{1}{n} \sum_{k=1}^n \frac{1}{\left(\sqrt{\frac{k-1/2}{n}}+\sqrt{\frac{k-1/2}{n}}\right)}\) By the Fundamental Theorem of calculus, the far left and far right hand sides of the inequality both approach \(\frac{1}{2}\int_0^1\frac{dx}{2\sqrt{x}}=\frac{1}{2}\) in the limit \(n\to\infty\), hence by applying the Squeeze Theorem, we get \(\lim_{n\to\infty}\frac{1}{\sqrt{n}}\sum_{k=1}^n (-1)^k\sqrt{k}=\frac{1}{2}\ \lim_{n\to\infty}\frac{1}{n} \sum_{k=1}^n \frac{1}{\left(\sqrt{\frac{k}{n}}+\sqrt{\frac{k}{n}-\frac{1}{2n}}\right)}=\frac{1}{2}\) All of the methods used above were available to me in high school, although as I mentioned I learned these things from independent reading (Schaum's Outlines Advanced Calculus, ancient version used by my dad back when he was a student). However, the high school curriculum itself didn't cover any of this stuff. It was an insult to mathematics that most of my classmates graduated from high school without ever having seen or understood a single proof relating to any of the core methods they were taught. Most of my math classmates still had the impression at the end that things like the Pythagorean theorem are proved by carefully constructing a wide variety of right triangles and measuring them, which would put them about 3000 years behind the times. Many were shocked when I mentioned you can calculate pi to any accuracy you want without actually drawing anything; they knew that more than 2000 years ago. :wallbang: rpenner, I've gone through your solutions and I have a few issues with them I'd like to ask you about later. In the meantime, I'd like to see if anyone can poke holes in my honest-to-goodness high-school-flavoured solution Please Register or Log in to view the hidden image!
Maybe I am missing something but you still have to show that the right hand side (upper bound) in the third displayed equation has limit that has a suitable dependence on epsilon.
Alrighty, I reworked the problematic portion of the proof, now there's no mention of any \(\epsilon\) whatsoever (which is good, because I just realized after your post that I was discussing square roots of negative numbers!!!). I think now the proof is complete, but if you still see problems let me know.
CptBork very closely repeated my own solution. With the exception of maybe the indefinite Riemann integral (which you can take an Eulerian approach to anyway because FToC works just fine) I'm sure you can now agree that 'clever' manipulations are the only thing needed to solve this. My first reaction was to go for Taylor series but I'm not too confident with my fluency in those.
In my solution though, I used a definite integral over the interval [0,1]. Both ends of my inequality are just slightly different tagged Riemann partitions for the same integral (when n is finite). So not quite sure what you mean by indefinite integrals, or the Eulerian approach. Anyhow like I said, none of my high school classmates were able to do anything remotely close to this, but I'm pretty sure our textbooks did give a brief overview of how Riemann sums and integrals were related, and I certainly knew about it. rpenner, in your first solution you replace a sum of discretely-spaced values with an integral over [0,n]. I know that wouldn't hold in general, but I guess since there's a \(1/\sqrt{n}\) factor on the outside, they must become the same in the asymptotic limit. What say you? Is that part of the reasoning when you introduce this integral?
I was referring to the \(x^{\frac{{ - 1}}{2}} \) integrand where the lower limit of integration is 0 and how that might be considered advanced.
Ahhhh, well in that case it's interesting that my dumbass high school calculus class actually spent a bit of time covering those types of integrals. I think the correct term is "improper integral", indefinite as I learned it comes when you don't specify limits on the integral. But I do see how giving a rigorous justification for Riemann's technique with improper integrals could be a bit tricky. Remember that most high school students can't rigorously justify derivatives and limits in general, though (we did learn a tiny bit about epsilon and delta in class, and I read a lot of that stuff on my own too out of Schaum's).
