Examine whether the series ∞∑ 1/(logn)^logn∑ is convergent.n=2

kuCAu 2021-08-18 Answered

Examine whether the series n=21(logn)logn is convergent. n=2

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Expert Answer

aprovard
Answered 2021-08-19 Author has 94 answers

Notice that, for all n2
(logn)logn=elog((logn)logn)=(elogn)log(logn)=nlog(logn)
Also note that log(logn)>2 for all n>ee2. Choose n0 such that n0>ee2. Then for all nn0 we have 1(logn)logn=1nlog(logn)1n2
The given series is convergent by comparison test.

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