Thread Closed

Convergence of a sequence

 
Share Thread Thread Tools
Oct15-04, 03:29 PM   #1
 
Recognitions:
Gold Membership Gold Member
Homework Helper Homework Help
Science Advisor Science Advisor

Convergence of a sequence


Apparently (according to my textbook), the sequence defined by

[tex]\left\{\frac{1}{n^2}+\frac{2}{n^2}+...+\frac{n-1}{n^2}\right\}[/tex]

converges towards 1/2, i.e. has 1/2 as a limit.

How could that be?! It seems to me that as n approaches infinity, all the fractions fall to zero. What is it I'm missing?
PhysOrg.com
PhysOrg
science news on PhysOrg.com

>> Galaxies fed by funnels of fuel
>> The better to see you with: Scientists build record-setting metamaterial flat lens
>> Google eyes emerging markets networks
Oct15-04, 03:44 PM   #2
 
Recognitions:
Gold Membership Gold Member
Science Advisor Science Advisor
Retired Staff Staff Emeritus
The question is if they tend to zero faster than their number grow towards infinity.
Oct15-04, 03:53 PM   #3
 
Recognitions:
Gold Membership Gold Member
Homework Helper Homework Help
Science Advisor Science Advisor


Is there a way to find this analytically?
Oct15-04, 04:06 PM   #4
 
Recognitions:
Gold Membership Gold Member
Homework Helper Homework Help
Science Advisor Science Advisor

Convergence of a sequence


Quote by quasar987


Is there a way to find this analytically?
Sure; you may write the partial sum as:
[tex]\frac{1}{n^{2}}(1+2+++n-1)=\frac{1}{n^{2}}\frac{n(n-1)}{2}[/tex]
Oct15-04, 05:21 PM   #5
 
Recognitions:
Gold Membership Gold Member
Homework Helper Homework Help
Science Advisor Science Advisor
Quote by arildno
Sure; you may write the partial sum as:
[tex]\frac{1}{n^{2}}(1+2+++n-1)=\frac{1}{n^{2}}\frac{n(n-1)}{2}[/tex]
Gauss's sum again! Damn! You guys are smart, are you all doctors in mathematics or physics?
Oct15-04, 06:26 PM   #6
 
Some of them are. :) Don't worry about it, I feel the same way you do all the time.
Oct15-04, 08:26 PM   #7
 
Recognitions:
Gold Membership Gold Member
Homework Helper Homework Help
Science Advisor Science Advisor
I posted a new thread in "College Level Help" by mistake: http://www.physicsforums.com/showthread.php?t=47964
Oct26-04, 06:32 PM   #8
 
Recognitions:
Gold Membership Gold Member
Homework Helper Homework Help
Science Advisor Science Advisor
Retired Staff Staff Emeritus
Quote by quasar987
Apparently (according to my textbook), the sequence defined by

[tex]\left\{\frac{1}{n^2}+\frac{2}{n^2}+...+\frac{n-1}{n^2}\right\}[/tex]

converges towards 1/2, i.e. has 1/2 as a limit.
Hold on a second. How is it that the index appears in every term when you list out the series?

Also, the above is a series, not a sequence.

How could that be?! It seems to me that as n approaches infinity, all the fractions fall to zero. What is it I'm missing?
The limit of the sequence is zero.
The limit of the sequence of partial sums is 1/2.
Oct26-04, 07:22 PM   #9
 
Recognitions:
Homework Helper Homework Help
Science Advisor Science Advisor
Quote by Tom Mattson
Hold on a second. How is it that the index appears in every term when you list out the series?

Also, the above is a series, not a sequence.
Tom, it's precisely the fact that the index "n" appears in each of the terms that makes this a sequence, and not a series, as it's given.

[tex]a_n=\sum_{i=1}^{n-1}\frac{i}{n^2}[/tex]

It's the limit of [tex]a_n[/tex] he's after. Since each of the terms in the sum is dependant on n, you can't break it into a series as I suspect you are thinking of doing.


You can of course think of any sequence as a series, by setting [tex]b_1=a_1, b_n=a_n-a_{n-1}[/tex], then [tex]a_n=\sum_{i=1}^{n}b_i[/tex], but that can be an awkward thing to do. In this case we'd find [tex]b_n=\frac{1}{2n(n+1)}[/tex], but I don't think that's what you were getting at?
Oct26-04, 07:52 PM   #10
 
Recognitions:
Gold Membership Gold Member
Homework Helper Homework Help
Science Advisor Science Advisor
Retired Staff Staff Emeritus
Do me a favor and just ignore me for the rest of the night....
Thread Closed
Thread Tools


Similar Threads for: Convergence of a sequence
Thread Forum Replies
convergence of a sequence Calculus & Beyond Homework 19
Convergence of a sequence General Math 2
Convergence of a sequence Calculus 4
Convergence of a Sequence Calculus & Beyond Homework 4
Isn't this weird? (convergence of a sequence) Introductory Physics Homework 5