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need help with math1241 questions (1 Viewer)

brett86

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determine the convergence or divergence of Σa<sub>n</sub> for the following a<sub>n</sub>:



thanks in adv to anyone that can help
 

withoutaface

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Well the second one diverges because its terms are all > than those of the harmonic series (1/n) which we know diverges.
 

Li0n

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waf..
do you really have a girlfriend?
but yeah.. the imaginary one is she really hotter than dreamerish :O:O
 

MAICHI

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The first one you would argue that (log(n))^3 < (n^(1/9))^3 = n^(1/3) for large n, now since n^(1/3) as the numerator converges so does the series.

The second one you would do a comparison test with 1/n, eg [1/n^(1+1/n)]/[1/n] = 1 as n approaches infinity, and we know that if one series diverges both series diverges, so the original series diverges since 1/n diverges.

The third one is just like the first one I think, log(n!) < n^(3/2) for large n.
 

gman03

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the third one use the fact that log(n!) < log(nn) = n log n
 

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