Answer
Divergent
Work Step by Step
Let us consider $f(x)=\dfrac{x}{x^2+4}$
Here, the function $f(x)$ is positive, continuous and decreasing for $x \geq 1$
Then $\int_3^\infty \dfrac{x}{x^2+4}dx= \lim\limits_{k \to \infty} \int_3^k \dfrac{x}{x^2+4}dx= \lim\limits_{k \to \infty} [(\dfrac{1}{2})\ln (x^2+4)]_3^k$
and $\lim\limits_{k \to \infty} [(\dfrac{1}{2}) \ln (k^2+4)-(\dfrac{1}{2}) \ln 13]= \infty$
Thus, the sequence $\Sigma_{n=1}^\infty \dfrac{n}{n^2+4}$ is Divergent.
