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?
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