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Mathematics
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Integral and Comparison Problem
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The first part seems to be convergent by the geometric series?
compare to 1/3^n = n/n*3^3>n/(n+1)*3^n for the first
Yes so the first one is convergent.
the second also converges. compare to n/n^3 > n/n^3-2n+6 for n > 3
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Could I compare the second one with 1/n^2?
Alrighty, how would I do the last one?
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
i think it's still gonna diverge but you'll have to check
Okay thank you!
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you're welcome
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