Answer
Conditionally convergent
Work Step by Step
Consider $u_n=\dfrac{1}{1+\sqrt n}$
Here, $u_n$ is positive for all the values of $n$.
Now, $f(n)=\dfrac{1}{1+\sqrt n}$
and $f'(n)=\dfrac{-1}{2\sqrt {n}(1+\sqrt n)^2} \lt 0$
The negative sign shows that, the sequence $u_n$ is not increasing.
Thus, $\lim\limits_{n \to \infty} u_n=\lim\limits_{n \to \infty}\dfrac{1}{1+\sqrt n}=0$
Thus, by the Alternating Series Test, the series converges.
Since, we know that $\Sigma_{n=1}^\infty \dfrac{1}{\sqrt n}=\Sigma_{n=1}^\infty \dfrac{1}{n^{1/2}}$, which a divergent p-series with $p=\dfrac{1}{2}$
Thus, the series does not converge absolutely. Hence, the series is conditionally convergent.