My Math Forum  

Go Back   My Math Forum > College Math Forum > Real Analysis

Real Analysis Real Analysis Math Forum


Thanks Tree3Thanks
  • 1 Post By Integrator
  • 1 Post By Maschke
  • 1 Post By SDK
Reply
 
LinkBack Thread Tools Display Modes
June 14th, 2019, 07:05 AM   #1
Member
 
Joined: Aug 2018
From: România

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
Integrator is offline  
 
June 16th, 2019, 09:12 PM   #2
Member
 
Joined: Aug 2018
From: România

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
Thanks from idontknow
Integrator is offline  
June 17th, 2019, 12:20 PM   #3
Math Team
 
topsquark's Avatar
 
Joined: May 2013
From: The Astral plane

Posts: 2,226
Thanks: 908

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
Originally Posted by Integrator View Post
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
topsquark is offline  
June 17th, 2019, 08:31 PM   #4
Member
 
Joined: Aug 2018
From: România

Posts: 84
Thanks: 6

Quote:
Originally Posted by topsquark View Post
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
Integrator is offline  
June 17th, 2019, 09:54 PM   #5
Member
 
Greens's Avatar
 
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.
Greens is offline  
June 24th, 2019, 09:33 PM   #6
Member
 
Joined: Aug 2018
From: România

Posts: 84
Thanks: 6

Hello all,

Some say the limit is $\displaystyle -\frac{1}{4\pi}$...

All the best,

Integrator
Integrator is offline  
June 24th, 2019, 10:27 PM   #7
Senior Member
 
Joined: Aug 2012

Posts: 2,342
Thanks: 731

Quote:
Originally Posted by Integrator View Post
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?
Thanks from Greens
Maschke is offline  
June 24th, 2019, 10:53 PM   #8
Member
 
Greens's Avatar
 
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.
Greens is offline  
June 24th, 2019, 11:31 PM   #9
Senior Member
 
Joined: Aug 2012

Posts: 2,342
Thanks: 731

Quote:
Originally Posted by Greens View Post
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.
Maschke is offline  
July 6th, 2019, 02:11 PM   #10
SDK
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.
Thanks from topsquark

Last edited by SDK; July 6th, 2019 at 02:24 PM.
SDK is offline  
Reply

  My Math Forum > College Math Forum > Real Analysis

Tags
limit



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
upper limit = lower limit implies convergence zylo Calculus 13 May 31st, 2017 12:53 PM
Limit jones123 Calculus 3 December 8th, 2012 02:27 PM
Limit Superior and Limit Inferior veronicak5678 Real Analysis 4 August 22nd, 2011 10:07 AM
Limit ..... Calculus 1 April 21st, 2011 09:15 AM
when should we evaluate left limit and right limit? conjecture Calculus 1 July 24th, 2008 01:14 PM





Copyright © 2019 My Math Forum. All rights reserved.