I'm trying to find out now whether the series
∑∞n=2an converges or not when
an=nlogn(logn)n
Again, I tried d'Alembert an+1an, Cauchy condensation test ∞∑n=22na2n, and they both didn't work for me.
I can't use Stirling, nor the integral test.
Edit: I'm searching for a solution which uses sequences theorems and doesn't involve functions.
Thank you
Answer
A comparison test will work here; the key is to write both numerator and denominator in terms of exponentials with bases not involving n. Note that the numerator is elog2n, which is less than en/2 for sufficiently large n. The denominator is enloglogn, which is greater than en for sufficiently large n; so for all sufficiently large n the terms are less than e−n/2 and thus the series converges.
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