# Checking the convergence of a series

this is one of many series that I have to check whether it converges or diverges. and to also check absolute convergence if possible:

∑e3√nnlog n+1∑en√3nlog n+1\sum{\frac{e^\sqrt[3]{n}}{n^{\log ~ n} + 1}}

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Is lognlogn\log n at the exponent?
– zar
Oct 20 at 22:51

yes that’s what I meant
– HexaFlexagon
Oct 20 at 22:51

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1

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Edit: We know that nlog(n)=elog2(n)nlog(n)=elog2(n)n^{\log(n)}=e^{\log^2(n)}. Hence:
e3√nnlog(n)+1=en1/3elog2(n)+1=en1/3−log2(n)11+e−log2(n)⟶∞en√3nlog(n)+1=en1/3elog2(n)+1=en1/3−log2(n)11+e−log2(n)⟶∞\dfrac{e^{\sqrt[3]{n}}}{n^{\log(n)}+1}=\dfrac{e^{n^{1/3}}}{e^{\log^2(n)}+1}= e^{n^{1/3}-\log^2(n)}\dfrac{1}{1+e^{-\log^2(n)}}\longrightarrow\infty
as n→∞n\rightarrow\infty. The series must diverge since the general term tends to infinity.

+1 i was about to write the same
– El Bazzi
Oct 20 at 22:50

i’m sorry if my notation was bad,
– HexaFlexagon
Oct 20 at 22:50

but it’s n^log n
– HexaFlexagon
Oct 20 at 22:50