Wednesday, 29 June 2016

real analysis - Studying the character of sumin=3nftyfrac1n(log(logn))alpha




I have to study the character of this series
n=31n(log(logn))α



with α a real parameter.



Considering the Cauchy condensation test, the equivalent series is:
n=12n12n[log(log2n)]α=n=11[log(nlog2)]α



Using the ratio test:




lim



Then when (-0,36)^{\alpha}>1 the given series converges, otherwise it diverges



if \alpha =1 , -0,36<1 , diverges



if \alpha =0 , 1<1 , diverges



if \alpha =-1 , (-0,36)^{-1}=-2,77<1 , converges




if \alpha >1 , (-0,36)^{\alpha}<1 , converges



if \alpha <-1 , (-0,36)^{\alpha}<1 , converges



if |\alpha| <1 , \ne 0 sometimes it doen't exist (-0,36)^{\alpha}
Can someone help me?


Answer



HINT




Let use limit comparison test with



\sum \frac{1}{n}


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