Answer
Diverges
Work Step by Step
Given $$ \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}+\ln n}$$
Compare with the divergent series $\displaystyle \sum_{n=1}^{\infty}\frac{1}{\sqrt{n} }$ ( $p-$ series , p<1) and by using the Limit Comparison Test, we get:
\begin{align*}
\lim_{n\to \infty} \frac{a_n}{b_n}&=\lim_{n\to \infty}\frac{\sqrt{n}}{\sqrt{n}+\ln n}\ \ \cr & \text{Using l'Hopital's rule}\\
&=\lim_{n\to \infty}\frac{\frac{1}{2 \sqrt{n}}}{\frac{1}{2 \sqrt{n}}+\frac{1}{n}}\\
&=1
\end{align*}
Then $\displaystyle \sum_{n=1}^{\infty}\frac{1}{\sqrt{n}+\ln n}$ also diverges
