Answer
Divergent
Work Step by Step
Consider $f(x)=\dfrac{x}{x^2+4}$ This is positive, continuous and decreasing for $x \geq 1$
Now, take the integral test to find the convergence and divergence for the sequence.
We have $\int_3^\infty \dfrac{x}{x^2+4}dx= \lim\limits_{a \to \infty} \int_3^a \dfrac{x}{x^2+4}dx$
or, $ \lim\limits_{a \to \infty} [\dfrac{1}{2}\ln (x^2+4)]_3^a=\lim\limits_{a \to \infty} [\dfrac{1}{2} \ln (a^2+4)-\dfrac{1}{2} \ln 13]= \infty$
Hence, the sequence $\Sigma_{n=1}^\infty \dfrac{n}{n^2+4}$ is Divergent.
