Integral Comparison Test
Compare improper Riemann integrals to infinite series.
Discover Bridges
Study Integral Comparison Test
Now we use integrals on unbounded domains to check whether series are convergent or not.
Theorem 1 (Integral Criterion for Series). Let \(f:[0,\infty)\to\mathbb{R}\) be monotonically decreasing and non-negative. Then the series \[\sum_{k=0}^{\infty}f(k)\] is convergent if and only if the integral \[\int_{0}^{\infty}f(x)dx\] converges. In the case of convergence, the following estimate holds true: \[0\leq \sum_{k=0}^\infty f(k) -\int_0^\infty f(x)~dx\leq f(0)\]
Proof: Monotonicity of \(f\) implies that for \(k-1\leq x\leq k\) holds \(f(k)\leq f(x)\leq f(k-1)\). Monotonicity of the integral therefore leads to the inequality \[f(k)=\int_{k-1}^kf(k)dx\leq \int_{k-1}^kf(x)dx\leq \int_{k-1}^kf(k-1)dx=f(k-1).\] Therefore \[\int_1^{n+2}f(x)dx\leq\sum_{k=1}^{n+1} f(k)\leq \int_0^{n+1}f(x)dx\leq \sum_{k=0}^{n} f(k).\] Using this inequality, we can directly conclude that, if one of the limits (integral or sum) as \(n\to\infty\) exists, then the other limit (sum or integral) also exists.
In case of convergence necessarily \((f(k))_{k\in\mathbb{N}_0}\) is a zero sequence. Therefore \[0\leq \sum_{k=0}^n f(k) -\int_0^{n+1} f(x)~dx\leq \sum_{k=0}^n f(k) - \sum_{k=1}^{n+1}f(k) =f(0)-f(n+1)\] implies the inequality in Theorem 1 for \(n\rightarrow \infty\). \(\Box\)
Remark 2. Since the convergence of \(\int_{a}^{\infty}f(x)dx\) does not depend on \(a\in\mathbb{R}\), the above result can also be slightly generalised in a way that the convergence of the series \(\sum_{k=a}^{\infty}f(k)\) for \(a\in\mathbb{N}\) is equivalent to that of the integral \(\int_{a}^{\infty}f(x)dx\).
Example 3.
Using the inproper integral of the function \(\int_0^\infty \frac{1}{x^\alpha} dx\), we see that \[\sum_{k=1}^\infty\frac1{k^\alpha}\] converges for \(\alpha>1\) and is divergent for \(\alpha\leq1\).
For \(\alpha\in\mathbb{R}\) consider \[\sum_{n=2}^\infty\frac1{n(\log(n))^\alpha}.\] We have \[\begin{aligned} \int_2^\infty\frac1{x(\log(x))^\alpha}~dx=&\lim_{n\to\infty}\begin{cases}\left.\frac{(\log(x))^{1-\alpha}}{1-\alpha}\right|_{x=2}^{x=n}&:\alpha\neq1,\\ \left.\log(\log(x))\right|_{x=2}^{x=n}&:\alpha=1\end{cases}\\ =&\begin{cases}-\frac{(\log(2))^{1-\alpha}}{1-\alpha}&:\alpha>1,\\ \infty&:\alpha\leq1.\end{cases} \end{aligned}\] Therefore, the series \[\sum_{n=2}^\infty\frac1{n(\log(n))^\alpha}.\] is convergent if and only if \(\alpha>1\).
Discuss your questions by typing below.
Solve the Exercise
The data for the interactive network on this webpage was generated with pntfx Copyright Fabian Gabel and Julian Großmann. pntfx is licensed under the MIT license. Visualization of the network uses the open-source graph theory library Cytoscape.js licensed under the MIT license.