Answer
Both the series diverges.
Work Step by Step
a) We will apply the limit comparison test with $a_n=\Sigma_{n=1}^{\infty} \dfrac{\ln n}{n}$ and $b_n=\Sigma_{n=1}^{\infty} \dfrac{1}{n}$
We can see that the series $\Sigma b_n$ shows a divergent p-series with $p=1$.
Next, we have $\lim\limits_{n \to \infty} \dfrac{a_n}{b_n} =\lim\limits_{n \to \infty} [\dfrac{\dfrac{\ln n}{n}}{1/n}]$
or, $=\lim\limits_{n \to \infty} \ln (n)$
or, $=\infty$
b) We will apply the limit comparison test with $a_n=\Sigma_{n=1}^{\infty} \dfrac{1}{\ln n}$ and $b_n=\Sigma_{n=1}^{\infty} \dfrac{1}{n}$
We can see that the series $\Sigma b_n$ shows a divergent p-series with $p=1$.
Next, we have $\lim\limits_{n \to \infty} \dfrac{a_n}{b_n} =\lim\limits_{n \to \infty} [\dfrac{\dfrac{1}{\ln n}}{1/n}]$
or, $=\lim\limits_{n \to \infty}\dfrac{ \ln (n)}{n}$
or, $=\infty$
Hence, we can see that both the series diverges by the limit comparison test.
