Using the Integral Test

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.