Sequence Convergence/Divergence Question

  • #1

Homework Statement


Determine which of the sequences converge or diverge. Find the limit of the convergent sequences.

1) {asubn}= [((n^2) + (-1)^n)] / [(4n^2)]



Homework Equations


[/B]
a1=first term, a2=second term...an= nth term

The Attempt at a Solution


a) So I found the first couple of terms
a1=0
a2=5/16
a3=2/9
a4=17/64

It didn't really look like it converged, but I took the limit as n approached infinity, divided the highest degree in the numerator and the denominator and got the limit to be 1/4. [/B]
 

Answers and Replies

  • #2
Math_QED
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Well, what exactly is the problem? The limit of this sequence is 1/4, as you have found yourself. I think you made a mistake in your thought proces though.

Lim(n->∞) [((n^2) + (-1)^n)] / [(4n^2)] = Lim(n->∞) n^2/(4n^2) + Lim(n->∞) (-1)^n/(4n^2) = 1/4 + Lim(n->∞) (-1)^n/(4n^2)
The second of those 2 limits goes to zero, because the numerator (-1)^n = -1, 1, -1, 1, -1, ... and the denominator becomes larger and larger. Only taking the highest degree in numerator and denumerator is not enough because (-1)^n has n in the exponent!
 
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  • #3
So because the denominator shoots off into infinity and also because (-1)^n oscillates between -1 and 1 forever, the sequence is divergent?
 
  • #4
Math_QED
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So because the denominator shoots off into infinity and also because (-1)^n oscillates between -1 and 1 forever, the sequence is divergent?
No I meant that Lim(n->∞) (-1)^n/(4n^2) = 0 and therefor: Lim(n->∞) [((n^2) + (-1)^n)] / [(4n^2)] = Lim(n->∞) n^2/(4n^2) + Lim(n->∞) (-1)^n/(4n^2) = 1/4 + 0 = 1/4, so the limit is 1/4.

From what I understand, you found the limit by taking the highest degree in numerator and denumerator, but since there is an n in the exponent, you have to be careful.
 
  • #5
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Homework Statement


Determine which of the sequences converge or diverge. Find the limit of the convergent sequences.

1) {asubn}= [((n^2) + (-1)^n)] / [(4n^2)]



Homework Equations


[/B]
a1=first term, a2=second term...an= nth term

The Attempt at a Solution


a) So I found the first couple of terms
a1=0
a2=5/16
a3=2/9
a4=17/64

It didn't really look like it converged, but I took the limit as n approached infinity, divided the highest degree in the numerator and the denominator and got the limit to be 1/4. [/B]
The dominant term in the numerator is n2, and the dominant term (and only term) in the denominator is 4n2. By "dominant" I mean that the larger n gets, the more insignificant the (-1)n term is. As you say, this sequence converges to 1/4.

You can split this sequence into two sequences: ##\{\frac{n^2}{4n^2} \}## and ##\{\frac{(-1)^n}{4n^2} \}##. If you can convince yourself that both sequences converge, then the sum of these two sequences will also converge.
 
  • #6
Ray Vickson
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So because the denominator shoots off into infinity and also because (-1)^n oscillates between -1 and 1 forever, the sequence is divergent?
The sequence ##s_n = (-1)^n/n^2## has ##s_{10} = 0.01##, ##s_{20} = 0.0025##, ##s_{1000}= 10^{-6}##, ##s_{1000000} = 10^{-12}##, ##s_{1000001}= -.999998 \, 10^{-12}##, etc. Does that look to you like a divergent sequence to you?
 
  • #7
The sequence ##s_n = (-1)^n/n^2## has ##s_{10} = 0.01##, ##s_{20} = 0.0025##, ##s_{1000}= 10^{-6}##, ##s_{1000000} = 10^{-12}##, ##s_{1000001}= -.999998 \, 10^{-12}##, etc. Does that look to you like a divergent sequence to you?
No need to be condescending, sir.
 
  • #8
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So because the denominator shoots off into infinity and also because (-1)^n oscillates between -1 and 1 forever, the sequence is divergent?
The sequence ##s_n = (-1)^n/n^2## has ##s_{10} = 0.01##, ##s_{20} = 0.0025##, ##s_{1000}= 10^{-6}##, ##s_{1000000} = 10^{-12}##, ##s_{1000001}= -.999998 \, 10^{-12}##, etc. Does that look to you like a divergent sequence to you?
No need to be condescending, sir.
I don't see Ray's response as being condescending. He was responding directly to something you said.
 

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