Answer
convergent
Work Step by Step
Given $$\sum_{n=1}^{\infty} \frac{1 }{n\sqrt{n^2+1}}$$
Use the Limit Comparison Test with $a_n =\dfrac{1 }{n\sqrt{n^2+1}}$ and $b_n=\dfrac{ 1}{ n^2}$
\begin{align*}
\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}&=\lim _{n \rightarrow \infty} \frac{n^2 }{n\sqrt{n^2+1}}\\
&=\lim _{n \rightarrow \infty} \frac{n/n }{ \sqrt{n^2/n^2+1/n^2}}\\
&=\lim _{n \rightarrow \infty}\frac{1 }{ \sqrt{1+1/n^2}} \\
&=1
\end{align*}
Since $\displaystyle \sum_{n=1}^{\infty} \frac{ 1 }{ n^2}$ is convergent ($p-$series series $p>1$) , then $\displaystyle\sum_{n=1}^{\infty} \frac{1 }{n\sqrt{n^2+1}}$ is also convergent