Divergent Series - How to test it

  • Thread starter Saterial
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  • #1
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Homework Statement



Ʃ (1+n+n2)/(√1+n2+n6)) n=1 to infinity
Answer is divergent


Homework Equations


Comparison Test / P Series
0≤an≤bn

The Attempt at a Solution


Hello, I simplified the problem to

Ʃ (1+n+n2)/(1+n2+n6)1/2

Is it incorrect of me to say immediately right here that because the power of the denominator p<1 it is divergent?

Or if I take a different approach use comparison test and say 0≤an≤1/(1+n2+n6)1/2 can I say it is divergent because bn is divergent due to p-series?

WA says it is divergent due to comparison test but how would I get something like say 1/n^6 out of that square root to use as bn?
 

Answers and Replies

  • #2
Simon Bridge
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I would have said it was divergent because it behaves as 1/n for large n... and we know the sum of 1/n diverges from a previous proof.
 
  • #3
Dick
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I would have said it was divergent because it behaves as 1/n for large n... and we know the sum of 1/n diverges from a previous proof.
True, but 1/n is on the boundary between divergence and convergence for power series. You need an actual comparison to make. Being asymptotically 'like' 1/n is not good enough. Choose a series with a smaller denominator and a larger denominator and show it diverges. I'm a little worried that what Saterial has said so far makes little sense.
 
  • #4
LCKurtz
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Homework Statement



Ʃ (1+n+n2)/(√1+n2+n6)) n=1 to infinity
Answer is divergent


Homework Equations


Comparison Test / P Series
0≤an≤bn

The Attempt at a Solution


Hello, I simplified the problem to

Ʃ (1+n+n2)/(1+n2+n6)1/2

Is it incorrect of me to say immediately right here that because the power of the denominator p<1 it is divergent?
Yes, that would be incorrect.

Or if I take a different approach use comparison test and say 0≤an≤1/(1+n2+n6)1/2 can I say it is divergent because bn is divergent due to p-series?
No. It isn't a p series

WA says it is divergent due to comparison test but how would I get something like say 1/n^6 out of that square root to use as bn?
Look up the "general comparison test" or it may be called the "limit comparison test" in your text.
 
  • #5
LCKurtz
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I would have said it was divergent because it behaves as 1/n for large n... and we know the sum of 1/n diverges from a previous proof.
That would be a correct intuitive argument, but you need the limit comparison test to prove it.
 
  • #6
Dick
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That would be a correct intuitive argument, but you need the limit comparison test to prove it.
A direct comparison test will do it. And it's simpler.
 
  • #7
LCKurtz
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A direct comparison test will do it. And it's simpler.
Probably so and point taken. I haven't actually worked on it. But I do think it is good for students to recognize situations that are a natural setting for the limit comparison test.
 
  • #8
Dick
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Probably so and point taken. I haven't actually worked on it. But I do think it is good for students to recognize situations that are a natural setting for the limit comparison test.
I'll repeat what I said before. Make the denominator smaller and the numerator larger in a way that the resulting series still diverges and you've got it. Limit comparison test may not be in Saterial's repetoire. And it's not needed. It's really simple once you've seen it.
 
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  • #9
Curious3141
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I'll repeat what I said before. Make the denominator smaller and the numerator larger in a way that the resulting series still diverges and you've got it. Limit comparison test may not be in Saterial's repetoire. And it's not needed. It's really simple once you've seen it.
Making the denominator smaller and the numerator larger (and hence every term larger) would not help you prove divergence.

Maybe you meant make the denominator larger and the numerator smaller, and prove that the resulting series still diverges?
 
  • #10
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Maybe you meant make the denominator larger and the numerator smaller, and prove that the resulting series still diverges?
Yes, the divergence follows rather trivially when one does that because the series collapses into the harmonic series.
 
  • #11
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(1+n+n2)/(√1+n2+n6)) > (1)/(√1+n2+n6)) > (1)/(√n2+n2+n6)) = (1)/(√2n2+n6)) = (1)/(√n2(2+n4))) = (1)/(n√(2+n4))) > 1/n
 
  • #12
Dick
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Making the denominator smaller and the numerator larger (and hence every term larger) would not help you prove divergence.

Maybe you meant make the denominator larger and the numerator smaller, and prove that the resulting series still diverges?
Oh yeah. That's what I meant. Denominator is the thing on the BOTTOM. Duh. Thanks!
 
  • #13
Dick
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(1+n+n2)/(√1+n2+n6)) > (1)/(√1+n2+n6)) > (1)/(√n2+n2+n6)) = (1)/(√2n2+n6)) = (1)/(√n2(2+n4))) = (1)/(n√(2+n4))) > 1/n
Sorry. (1)/(n√(2+n4))) is less than 1/n. Not greater.
 
  • #14
Dick
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How about just noting that the original series is greater than n^2/sqrt(n^6+n^6+n^6)?
 
  • #15
Curious3141
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How about just noting that the original series is greater than n^2/sqrt(n^6+n^6+n^6)?
Yes, that would work. Much simpler, too. :approve:

I just showed the first thing I thought of.
 
  • #16
uart
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Or as another alternative noting that it's greater than [itex]\frac{n^2}{n^3 + 1}[/itex], which in turn is greater than [itex]\frac{1}{n+1}[/itex]. We can then compare it directly to the harmonic series starting at n=2.
 

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