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 June 14th, 2019, 07:05 AM #1 Member   Joined: Aug 2018 From: RomÃ¢nia Posts: 88 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: 88 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 Thanks from idontknow
June 17th, 2019, 12:20 PM   #3
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 Originally Posted by Integrator 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
What limit? Is it $\displaystyle \lim_{n \to \infty} x_n$ or possibly $\displaystyle \lim_{n \to \infty} \sum _{n = 0}^{\infty} x_n$?

-Dan

June 17th, 2019, 08:31 PM   #4
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 Originally Posted by topsquark What limit? Is it $\displaystyle \lim_{n \to \infty} x_n$ or possibly $\displaystyle \lim_{n \to \infty} \sum _{n = 0}^{\infty} x_n$? -Dan
Hello,

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: 90 Thanks: 61 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: 88 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
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Quote:
 Originally Posted by Integrator 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 $\displaystyle n\in\mathbb{N}^{*}$.
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: 90 Thanks: 61 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
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Quote:
 Originally Posted by Greens 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.
Oh I see, quite right. Not all the $sin$ terms are zero.

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: 642 Thanks: 406 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. Thanks from topsquark Last edited by SDK; July 6th, 2019 at 02:24 PM.

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