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danmac0710
 3 years ago
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?
danmac0710
 3 years ago
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?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sig(n/an2+3)>sig(n/2n2+4n)=sig(1/n+2)=diverge >sig(n/2n2+3)=diverge

danmac0710
 3 years ago
Best ResponseYou've already chosen the best response.0I am sorry but I did not understand fully. Could you give more detail?

danmac0710
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks for helping me!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Use limit comparison test with divergent pseries 1/ndw:1356516645558:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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 ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0divide by 1/2n instead, it's your question asking for that

danmac0710
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks very much  just having a think!

danmac0710
 3 years ago
Best ResponseYou've already chosen the best response.0If 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.

danmac0710
 3 years ago
Best ResponseYou've already chosen the best response.0My textbook defines the Comparison Test this way anyhow.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1356517532318:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0in this case, your b series is 1/2n, which is diverge.dw:1356517694459:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The definition in my textbook is just simple as that. you get it?

danmac0710
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks 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!
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