% This is lecture no. 17 of MH1101/: Calculus II. \chapter{The Integral Test}% \label{cha:the_integral_test} \section{Intuition Behind Integral Test}% \label{sec:intuition_behind_integral_test} The \textbf{integral test} is a type of convergence test. A convergence test is a test we can attempt to apply on a series to better determine its convergence. We see how the test works using an example. \begin{exmp} Consider the series: \[ \sum_{n=1}^{\infty} \frac{1}{n^2} .\] It is known that the improper integral $\displaystyle{\int_{1}^{\infty} \frac{1}{x^2} \ud x}$ is convergent since: \[ \int_{1}^{\infty} \frac{1}{x^2} \ud x = \lim_{t \to \infty}\left[\frac{x^{-1}}{-1}\right]^{t}_{1} = 1 .\] Yet, we note that: \[ \frac{1}{2^2}+\frac{1}{3^2} + \dotsb + \frac{1}{n^2}\leqslant \int_{1}^{\infty} \frac{1}{x^2} \ud x = 1 .\] As seen in the following diagram: \begin{center} \begin{tikzpicture} \begin{axis}[ axis lines = middle, xlabel = {$x$}, ylabel = {$y$}, x = 2cm, y = 2cm, xmin = 0, xmax = 5.5, ymin = 0, ymax = 2] \addplot [ name path = A, domain = 0.8:5.5, samples = 100 ] {x^(-2)} node [pos=0, above] {$y=\ds{\frac{1}{x^2}}$}; \addplot [ name path = B, domain = 0:1 ] {1}; \addplot [ name path = C, domain = 0:1 ] {0}; \addplot [gray!30] fill between [of = B and C, soft clip={domain=0:1}]; \draw (axis cs:{0,0}) -- (axis cs:{0,1}); \draw (axis cs:{1,0}) -- (axis cs:{1,1}); \addplot [ name path = D, domain = 1:2 ] {0.25}; \addplot [ name path = E, domain = 1:2 ] {0}; \addplot [gray!30] fill between [of = D and E, soft clip={domain=1:2}]; \draw (axis cs:{2,0}) -- (axis cs:{2,0.25}); \addplot [ name path = F, domain = 2:3 ] {1/9}; \addplot [ name path = G, domain = 2:3 ] {0}; \addplot [gray!30] fill between [of = F and G, soft clip={domain=2:3}]; \draw (axis cs:{3,0}) -- (axis cs:{3,1/9}); \addplot [ name path = H, domain = 3:4 ] {1/16}; \addplot [ name path = I, domain = 3:4 ] {0}; \addplot [gray!30] fill between [of = H and I, soft clip={domain=3:4}]; \draw (axis cs:{4,0}) -- (axis cs:{4,1/16}); \addplot [ name path = J, domain = 4:5 ] {1/25}; \addplot [ name path = K, domain = 4:5 ] {0}; \addplot [gray!30] fill between [of = J and K, soft clip={domain=4:5}]; \draw (axis cs:{5,0}) -- (axis cs:{5,1/25}); \end{axis} \end{tikzpicture} \end{center} Therefore, the $k$-th partial sums of the series are bounded: \[ s_k = \frac{1}{1^2}+\frac{1}{2^2}+\dotsb+\frac{1}{k^2}\leqslant 1 + \int_{1}^{\infty} \frac{1}{x^2} \ud x = 1+ 1 = 2 .\] On top of that, the sequence of partial sums $\{s_k\}$ is increasing, since: \[ s_{k+1}=s_k + \frac{1}{(k+1)^2}>s_k .\] Hence, by the \textbf{monotonic convergence theorem}, the sequence $\{s_k\}$ is convergent, and thus the series is convergent. \end{exmp} \begin{thrm} The Integral Test. Suppose $f$ is a continuous, positive, decreasing function on $[1,\infty)$, and let $a_n = f(n)$. The following implications hold true: \begin{align*} \int_{1}^{\infty} f(x) \ud x\ \text{is convergent}\ & \implies \sum_{n=1}^{\infty} a_n\ \text{is convergent.}\\ \int_{1}^{\infty} f(x) \ud x\ \text{is divergent}\ & \implies \sum_{n=1}^{\infty} a_n\ \text{is divergent.} \end{align*} \end{thrm} \begin{note} When using the integral test, it is not necessary that the series or the integral starts at $n=1$. It is also not necessary that $f$ is always decreasing. It is necessary for $f$ to be \textit{ultimately} decreasing, that is, $f$ is decreasing $\forall x\geqslant N$, where $N\in\mathbb{Z}$. \end{note} \begin{exmp} Test the series $\displaystyle{\sum_{n=1}^{\infty} \frac{1}{1+n^2}}$ for convergence. \begin{sol} The function $f(x)= \displaystyle{\frac{1}{1+x^2}}$ is continuous, positive, and decreasing on $[1,\infty)$. $\therefore$ we can use the integral test: \begin{align*} \int_{1}^{\infty} \frac{1}{1+x^2} \ud x & = \lim_{t \to \infty}\int_{1}^{t} \frac{1}{1+x^2} \ud x\\ & = \lim_{t \to \infty}\left[\tan^{-1}x\right]^{t}_1\\ & = \lim_{t \to \infty}\left(\tan^{-1}t - \frac{\pi}{4}\right)\\ & = \frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}. \end{align*} Thus, by the integral test, the series $\displaystyle{\sum_{n=1}^{\infty} \frac{1}{1+n^2}}$ is convergent. \end{sol} \end{exmp} \section{$p$-series}% \label{sec:_p_series} An important class of series consists of $p$-series. The convergence of a $p$-series can be determined using the integral test. \begin{thrm} $p$-series. A $p$-series is a series of the form: \[ \sum_{n=1}^{\infty} \frac{1}{n^{p}} ,\] where $p$ is some fixed real number. It is convergent when $p>1$, and divergent when $p\leqslant 1$. \end{thrm} \begin{proof} We consider different cases depending on $p$. Say: \[ p<0 .\] Then the term $\displaystyle{\frac{1}{n^{p}}}$ can be arbitrarily large for sufficiently large $n$, that is: \[ \lim_{n \to \infty}\frac{1}{n^{p}}=\infty ,\] so, by the $n$-th term divergence test, the series is divergent. Now, consider the case where: \[ p=0 .\] In this case, the term $\displaystyle{\frac{1}{n^{p}}=1}$ for all values of $n$, and so by the $n$-th term divergence test, the series is divergent. For the case where: \[ p=1 ,\] we have $\displaystyle{\frac{1}{n^{p}}=\frac{1}{n}}$. Therefore, the $p$-series becomes the harmonic series which is known to be divergent. Finally, for the case where: \[ p\neq 1 \land p >0 ,\] we have: \begin{align*} \int_{1}^{\infty} \frac{1}{x^{p}} \ud x & = \lim_{t \to \infty}\int_{1}^{t} x^{-p} \ud x\\ & = \lim_{t \to \infty}\left[\frac{x^{-p+1}}{-p+1}\right]^{t}_{1}\\ & = \lim_{t \to \infty}\frac{1}{1-p}\left(t^{1-p}-1\right) \end{align*} Now, consider the case where $p>1$, then $p-1>0$, and so $t^{1-p}=\displaystyle{\frac{1}{t^{p-1}}}\to 0$ as $t\to\infty$. Therefore: \[ \int_{1}^{\infty} \frac{1}{x^{p}} \ud x = \lim_{t \to \infty}\frac{1}{1-p}\left(t^{1-p}-1\right) = \frac{1}{1-p}(0-1) = \frac{1}{p-1} .\] By the integral test, the corresponding $p$-series is convergent. Now, consider the case where $0
0$, and so $t^{1-p}\to\infty$ as $t\to\infty$. Therefore, \[ \int_{1}^{\infty} \frac{1}{x^{p}} \ud x = \lim_{t \to \infty}\frac{1}{1-p}\left(t^{1-p}-1\right)=\infty .\] By the integral test, the corresponding $p$-series is divergent. \end{proof} \begin{exmp} Determine whether $\displaystyle{\sum_{n=1}^{\infty} \frac{\ln n}{n}}$ is convergent. \begin{sol} Consider the function: \[ f(x)= \frac{\ln x}{x} .\] Clearly, $f(x)$ is positive and continuous for all $x>1$. To show that $f$ is decreasing, we compute the derivative: \[ f^{\prime}(x) = \frac{\left(\frac{1}{x}\right)x-\ln x}{x^2}= \frac{1-\ln x}{x^2} .\] Thus, $f^{\prime}(x)<0$ for all $\ln x > 1$, therefore $f$ is decreasing for all $x> e$ and $x\geqslant 3$. In view of the integral test, we check the improper integral: \begin{align*} \int_{3}^{\infty} \frac{\ln x}{x} \ud x & = \lim_{t \to \infty}\int_{3}^{t} \frac{\ln x}{x} \ud x\\ & = \lim_{t \to \infty}\left[\frac{1}{2}(\ln x)^2\right]^{t}_{3}\\ & = \lim_{t \to \infty}\left(\frac{(\ln t)^2}{2}-\frac{1}{2}(\ln 3)^2\right) = \infty. \end{align*} \end{sol} \end{exmp} By the integral test, the series is divergent.