June 14th, 2019, 07:05 AM  #1 
Member Joined: Aug 2018 From: RomÃ¢nia Posts: 84 Thanks: 6  A limit
Hello all, Find the limit of $\displaystyle x_{n}=\sum_{k=1}^{n}\bigg [\sqrt{n^{4}+k}\cdot\sin\bigg (\frac{2k\pi}{n}\bigg) \bigg ]$ where $\displaystyle n\in\mathbb{N}^{*}$. All the best, Integrator 
June 16th, 2019, 09:12 PM  #2 
Member Joined: Aug 2018 From: RomÃ¢nia Posts: 84 Thanks: 6 
Hello all, Some say the limit would be different from zero ...How do we calculate this limit? Thank you very much! Al the best, Integrator 
June 17th, 2019, 12:20 PM  #3  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,226 Thanks: 908 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
June 17th, 2019, 08:31 PM  #4  
Member Joined: Aug 2018 From: RomÃ¢nia Posts: 84 Thanks: 6  Quote:
What is $\displaystyle x_n$? I think it is clear that $\displaystyle x_n$ is the general term of a string...so we have to calculate the limit of that string. All the best, Integrator  
June 17th, 2019, 09:54 PM  #5 
Member Joined: Oct 2018 From: USA Posts: 87 Thanks: 59 Math Focus: Algebraic Geometry 
Yes, but limit to what? $\lim_{x \to 1}$ , $\lim_{x \to 5}$, $\lim_{x \to L}$ are all limits. Topsquark is asking what limit you're looking for. EDIT: I realized i've read incorrectly, sorry. Last edited by Greens; June 17th, 2019 at 09:56 PM. 
June 24th, 2019, 09:33 PM  #6 
Member Joined: Aug 2018 From: RomÃ¢nia Posts: 84 Thanks: 6 
Hello all, Some say the limit is $\displaystyle \frac{1}{4\pi}$... All the best, Integrator 
June 24th, 2019, 10:27 PM  #7 
Senior Member Joined: Aug 2012 Posts: 2,342 Thanks: 731  Am I missing something? $\sin 2 k \pi = 0$ for every integer $k$. Every term of the sum is zero. Each $x_n = 0$ and the limit (presumably as $n \to \infty$) is $0$. Yes? No? What am I missing?

June 24th, 2019, 10:53 PM  #8 
Member Joined: Oct 2018 From: USA Posts: 87 Thanks: 59 Math Focus: Algebraic Geometry 
I was thinking similarly, the $n$ in the denominator sends the sin to 0 regardless, but now I've got myself stuck thinking that the sin swaps sign at $k=\frac{n}{2}$.... but since the limit is n to infinity, n/2 is just infinity.....? I'll have to look more tomorrow, I'm falling asleep and the margins are too small. Last edited by skipjack; June 24th, 2019 at 11:22 PM. 
June 24th, 2019, 11:31 PM  #9  
Senior Member Joined: Aug 2012 Posts: 2,342 Thanks: 731  Quote:
It's late here too ... for some reason I was seeing it as the sin term then divided by n ... but of course that's not true. I wasn't seeing what I was missing and now I can't believe I saw what wasn't there. Time for bed. Last edited by Maschke; June 24th, 2019 at 11:37 PM.  
July 6th, 2019, 02:11 PM  #10 
Senior Member Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics 
Ahh this is cute but probably a bit too hard to give to a beginning calc student without some hints. 1. Define $2m = n$ and compute the limit for $m \to \infty$ instead (justify why its ok to compute this limit for only even $n$). 2. After using the odd symmetry of $\sin$ this makes the sum look like \[ \sum_{k = 1}^{n} \sqrt{n^4 + k} \sin \frac{2k \pi}{n} = \sum_{k = 1}^{m} \left(\sqrt{(2m)^4 + k}  \sqrt{(2m)^4 + m + k} \right) \sin \frac{k \pi}{m} \] 3. Do some mean value estimates on the term in parenthesis. As a hint, the mean value theorem gives you bounds similar to \[ \left(\sqrt{(2m)^4 + k}  \sqrt{(2m)^4 + m + k} \right) \approx \frac{1}{4m} \] for $m$ large enough. You can actually get explicit upper and lower bounds on this term so you can apply the squeeze theorem to your sum. 4. Returning to the sum you will just be left with showing that \[ \lim\limits_{m \to \infty} \frac{1}{4m} \sum_{k = 1}^m \sin \frac{k \pi}{m} = \frac{1}{4} \int_0^\pi \sin t \ dt \] which can be done by noticing that each of these sums is a Riemann sum. Last edited by SDK; July 6th, 2019 at 02:24 PM. 

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