Answer
converges to $\dfrac{\pi}{4}-\dfrac{1}{2}\tan^{-1}( \dfrac{1}{2})$
Work Step by Step
Consider $f(x)=\dfrac{1}{x^2+4}$. 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 $\int_1^\infty \dfrac{1}{x^2+4} dx= \lim\limits_{a \to \infty} \int_1^a \dfrac{1}{x^2+4} dx$
or, $ \lim\limits_{a \to \infty} [\frac{1}{2}\tan^{-1} \dfrac{x}{2}]_1^a=\lim\limits_{a \to \infty} [\frac{1}{2}\tan^{-1} \dfrac{a}{2}-\dfrac{1}{2}\tan^{-1} \dfrac{1}{2}]=\dfrac{\pi}{4}-\dfrac{1}{2}\tan^{-1}( \dfrac{1}{2})$
Hence, the sequence $\Sigma_{n=1}^\infty \dfrac{1}{n^{2}+4}$ converges to $\dfrac{\pi}{4}-\dfrac{1}{2}\tan^{-1}( \dfrac{1}{2})$
