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Mathematics 7 Online
OpenStudy (anonymous):

Integral and Comparison Problem

OpenStudy (anonymous):

OpenStudy (anonymous):

The first part seems to be convergent by the geometric series?

OpenStudy (anonymous):

compare to 1/3^n = n/n*3^3>n/(n+1)*3^n for the first

OpenStudy (anonymous):

Yes so the first one is convergent.

OpenStudy (anonymous):

the second also converges. compare to n/n^3 > n/n^3-2n+6 for n > 3

OpenStudy (anonymous):

Could I compare the second one with 1/n^2?

OpenStudy (anonymous):

Alrighty, how would I do the last one?

OpenStudy (anonymous):

i think the integral test. do a u sub with u = ln(n) and then do another u sub (actually do a t sub) with t = ln u and see what you get

OpenStudy (anonymous):

i think it's still gonna diverge but you'll have to check

OpenStudy (anonymous):

Okay thank you!

OpenStudy (anonymous):

you're welcome

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