Answer
Conditionally Convergent
Work Step by Step
Let us consider $a_n=\dfrac{1}{n+3}$
Here, $f(n)=\dfrac{1}{n+3} \implies f'(n)=\dfrac{-1}{(n+3)^2}\lt 0$
The negative sign shows that, the sequence $u_n$ is decreasing.
Then, $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \dfrac{1}{n+3}=0$
So, the series converges by the Alternating Test Series.
As we can see that $\Sigma_{n=1}^\infty \dfrac{1}{ n}$ diverges by the limit comparison test.
Therefore, the series do not converge absolutely this implies that the series is conditionally convergent.
