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- 590

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0 + ln(2)/4 + ln(3)/9 + ln(4)/16 + ...... + ln(n)/n^2

to find out if series converge/diverge, i have been using comparisons, finding a similar series, B(n) whose behaviour i know or can easily find, then saying :

# if B(n) > A(n), and B(n) converges, then A(n) converges

# if B(n) < A(n), and B(n) diverges, then A(n) diverges

# if A(n)/B(n)=K (K= not 0, not ∞) then A(n) and B(n) behave the same

but i cannot find any series that answers to any of these 3 condidtions, any ideas? the series must either have a known or relatively simple to find behaviour.