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As the title of this question states, how should I show $$\sum_{n=1}^\infty \frac{1}{1+n \ln(n)}$$ is divergent?

I'm thinking of using comparison test but cannot figure out an upper bound $g(n)$ for $$1+n\ln(n)$$ such that $\sum \frac{1}{g(n)}$ diverges.

I thought about using integral test but I think integrating $\frac{1}{1+n \ln(n)}$ is tedious...

Ratio test also does not seem to work.

Thanks

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Use the comparison test with the series$$\sum_{n=1}^\infty\frac1{n\log n},$$which diverges, by the integral test.

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By comparison test $\sum \frac 1{ 1+n\ln n}$ is convergent iff $\sum \frac 1{ n\ln n}$ is convergent. Now use the integral test.

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As an alternative by Cauchy condensation test consider

$$\sum_{n=1}^\infty 2^na_{2^n}=\sum_{n=1}^\infty \frac{2^n}{1+2^n \ln(2^n)}=\sum_{n=1}^\infty \frac{2^n}{1+n\cdot 2^n \ln 2}$$

which diverges by limit comparison test with $\sum \frac1n$, indeed

$$\frac{\frac{2^n}{1+n\cdot 2^n \ln 2}}{\frac1n}=\frac{n\cdot2^n}{1+n\cdot 2^n \ln 2}\to \frac1{\ln 2}$$

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    $\begingroup$ Should be $1+2^n\ln(2^n)$ in the denominator. $\endgroup$ – Milten Oct 28 '19 at 12:29
  • $\begingroup$ @Milten of course! Thanks $\endgroup$ – user Oct 28 '19 at 12:30

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