# How to show $\sum_{n=1}^\infty \frac{1}{1+n\ln(n)}$ is divergent?

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

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

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.

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

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

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