Answer
Conditionally convergent
Work Step by Step
Consider $a_n=\dfrac{1}{n+3}$
Here, $a_n$ is positive for all the values of $n$.
Let $f(n)=\dfrac{1}{n+3} \implies f'(n)=\dfrac{-1}{(n+3)^2}\lt 0$
The negative sign shows that, the sequence $u_n$ is not increasing.
Thus, $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \dfrac{1}{n+3}=0$
Thus, by the Alternating Test Series, the series converges.
We know that $\Sigma_{n=1}^\infty \dfrac{1}{ n}$ diverges.
Thus, the series does not converge absolutely.
Hence, the series is conditionally convergent.