Determine whether the series $\sum\limits_{k\,=\,1}^{\infty }a_{k}= \sum\limits_{k\,=\,1}^{\infty}\dfrac{4}{k^{2}+1}$ converges or diverges.

Solution The function $f(x)=\dfrac{4}{x^{2}+1}$ is defined on the interval $[1,\infty)$ and is continuous, positive, and decreasing for all numbers $x\geq 1$. Also, $a_{k}=f(k)$ for all positive integers $k.$ Using the Integral Test, we find $$\begin{eqnarray*} \int_{1}^{\infty }\!\dfrac{4}{x^{2}+1}\,dx\! : \lim_{b\,\rightarrow \,\infty }\int_{1}^{b}\dfrac{4}{x^{2}+1}\,dx&=&\lim_{b\,\rightarrow \,\infty } \,\left[ 4\int_{1}^{b}\dfrac{1}{x^{2}+1}\,dx\right] =4\lim_{b\,\rightarrow \,\infty }\,\left[\tan ^{-1}x\right] _{1}^{b} \\[12pt] &=& 4\lim_{b\,\rightarrow \,\infty }\,\left[\tan ^{-1}b-\tan ^{-1}1\right]\! =4\,\left[\dfrac{\pi }{2}-\dfrac{\pi }{4}\right]\! =\pi \end{eqnarray*}$$

Since the improper integral converges, the series $\sum\limits_{k\,=\,1}^{\infty}\dfrac{4}{k^{2}+1}$ converges.