## Thomas' Calculus 13th Edition

converges to $(\dfrac{\pi}{2})-\dfrac{1}{2}\tan^{-1}( \dfrac{1}{2})$
Let us consider $f(x)=\dfrac{1}{x^2+4}$. Here, the function $f(x)$ is positive, continuous and decreasing for $x \geq 1$ Then $\int_1^\infty \dfrac{1}{x^2+4} dx= \lim\limits_{k \to \infty} \int_1^k \dfrac{1}{x^2+4} dx=\lim\limits_{k \to \infty} [\dfrac{1}{2}\tan^{-1} \dfrac{x}{2}]_1^k$ and $\lim\limits_{k \to \infty} [(\dfrac{1}{2})\tan^{-1} (\dfrac{k}{2})-(\dfrac{1}{2})\tan^{-1} (\dfrac{1}{2})]=(\dfrac{\pi}{2})-(\dfrac{1}{2})\tan^{-1}( \dfrac{1}{2})$ Thus, the sequence $\Sigma_{n=1}^\infty \dfrac{1}{n^{2}+4}$ converges to $(\dfrac{\pi}{2})-\dfrac{1}{2}\tan^{-1}( \dfrac{1}{2})$.