Convergence of a sequence

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  • #1
quasar987
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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?
 

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  • #2
Hurkyl
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The question is if they tend to zero faster than their number grow towards infinity.
 
  • #3
quasar987
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:eek:

Is there a way to find this analytically?
 
  • #4
arildno
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quasar987 said:
:eek:

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]
 
  • #5
quasar987
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arildno said:
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?
 
  • #6
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Some of them are. :) Don't worry about it, I feel the same way you do all the time.
 
  • #8
Tom Mattson
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quasar987 said:
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.
 
  • #9
shmoe
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Tom Mattson said:
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?
 
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  • #10
Tom Mattson
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I really do know better than that...

Do me a favor and just ignore me for the rest of the night....
 
Last edited:

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