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danmac0710 Group Title

I need to show that this series diverges: n/(2n^2+3) by comparison with: 1/(2n). I think 1/(2n) diverges as it is simply 1/2 time the harmonic series. The problem I have is that my original series simplifies to: 1/(2n+3/n) once divided through by 'n' which is dominated by 1/(2n). I did not think it was possible to show a series diverges by showing a dominant series also diverges. Any advice please?

  • one year ago
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  1. mahmit2012 Group Title
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    sig(n/an2+3)>sig(n/2n2+4n)=sig(1/n+2)=diverge ->sig(n/2n2+3)=diverge

    • one year ago
  2. danmac0710 Group Title
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    I am sorry but I did not understand fully. Could you give more detail?

    • one year ago
  3. danmac0710 Group Title
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    Thanks for helping me!

    • one year ago
  4. hieuvo Group Title
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    Use limit comparison test with divergent p-series 1/n|dw:1356516645558:dw|

    • one year ago
  5. hieuvo Group Title
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    The trick is to make the power of denominator and numerator equal so that the limit exists, then the testing series and tested series both converge or diverge. U got it ?

    • one year ago
  6. hieuvo Group Title
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    divide by 1/2n instead, it's your question asking for that

    • one year ago
  7. danmac0710 Group Title
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    Thanks very much - just having a think!

    • one year ago
  8. danmac0710 Group Title
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    If I multiply my series by n/1 as you said, I then get a series which dominates my original series. I cannot show that my series diverges just by showing that a dominant series diverges, I need a series which mine dominates surely? Sorry if I am being stupid.

    • one year ago
  9. danmac0710 Group Title
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    My textbook defines the Comparison Test this way anyhow.

    • one year ago
  10. hieuvo Group Title
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    |dw:1356517532318:dw|

    • one year ago
  11. hieuvo Group Title
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    in this case, your b series is 1/2n, which is diverge.|dw:1356517694459:dw|

    • one year ago
  12. hieuvo Group Title
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    The definition in my textbook is just simple as that. you get it?

    • one year ago
  13. danmac0710 Group Title
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    Thanks very much - I had to go celebrate xmas yesterday so I logged off - I now understand your solution. The only things is that my text asked me to find the results using both the ratio comparison test which you helped me with in ADDITION to the other (not sure of the name?) comparison test which mean finding a series which dominates your series and converges or by finding a series which is dominated BY your series which diverges. It's abit like the Squeeze Theorem, but from one side only. It only works when all terms in both series are positive and the series is monotonic, i.e. always getting larger or smaller. Thanks very much - have a medal!

    • one year ago
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