The integral test

$\newenvironment {prompt}{}{} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[1]{{\overset {\rightharpoonup }{#1}}} \newcommand {\vec }[1]{{\overset {\boldsymbol {\rightharpoonup }}{\mathbf {#1}}}} \DeclareMathOperator {\proj }{\vec {proj}} \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\vec {\boldsymbol {\l }}} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} changed the theorem header of amsthm failed\MessageBreak }}} \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\Sectionformat }[2]{#1}$

Infinite sums can be studied using improper integrals.

We have seen that we can graph a sequence as a collection of points in the plane. For instance, consider the harmonic sequence, or the sequence where $a_n = 1/n$ : $1, \, \frac {1}{2}, \, \frac {1}{3}, \, \frac {1}{4}, \, \frac {1}{5}, \,\dots .$ If we plot the harmonic sequence, it looks like this.

Is there a nice way to visualize the sum $\sum _{k=1}^\infty \frac {1}{k}?$ One way to visualize the sum is to make rectangles whose areas are equal to the terms of the sequence.

The entire area above is exactly equal to the sum: $\sum _{k=1}^\infty \frac {1}{k}$ The previous image should remind you of a Riemann Sum, and for good reason. This technique lets us visually compare the sum of an infinite series to the value of an improper integral, as long as we imagine that we can draw infinitely many such rectangles. For instance, if we add a plot of $1/x$ to our picture above

we see that $\int _1^\infty \frac {1}{x} \d x < \sum _{k=1}^\infty \frac {1}{k}.$ This leads us to an interesting observation.

Let $f$ be continuous, positive, and decreasing on $[1,\infty )$ with $a_k = f(k)$ . If $\int _1^\infty f(x) \d x$ diverges, so does $\sum _{k=1}^\infty a_k$ .

That’s a pretty good observation, but we can do better. Consider the sequence $a_n = 1/n^2$ : $1, \, \frac {1}{4}, \, \frac {1}{9}, \, \frac {1}{16}, \, \frac {1}{25}, \,\dots .$ Think about it. If $\sum _{k=1}^\infty a_k$ diverges, then so does $\sum _{k=1}^\infty a_{k+1}.$ Why? The second sum is simply the first sum missing $a_1$ and no single number can be responsible for this sum diverging to infinity. This fact can be summed up in the following theorem:

Infinite Nature of Series The convergence or divergence of a series remains unchanged by the addition or subtraction of any finite number of terms. That is:

a divergent series will remain divergent with the addition or subtraction of any finite number of terms.
a convergent series will remain convergent with the addition or subtraction of any finite number of terms. (Of course, the sum will likely change.)

Back to the task at hand! Let’s graph $a_{k+1}$ using our rectangle technique.

The shaded area above is exactly equal to the sum $\sum _{k=1}^\infty \frac {1}{(k+1)^2}.$ We again visually compare the sum of an infinite series to the value of an improper integral. For instance, if we add a plot of $1/x^2$ to our picture above

we see that $\sum _{k=1}^\infty \frac {1}{k^2}. < \int _1^\infty \frac {1}{x^2} \d x.$ This conclusion leads us to another interesting observation.

Let $f$ be continuous, positive, and decreasing on $[1,\infty )$ with $a_k = f(k)$ . If $\sum _{k=1}^\infty a_k$ diverges, so does $\int _1^\infty f(x) \d x$ .

Together, these observations give us the integral test:

Integral Test Let $f$ be continuous, positive, and decreasing on $[1,\infty )$ with $a_k = f(k)$ for all $k > N$ for some index $N$ . Then, either $\sum _{k=1}^\infty a_k\text { and }\int _1^\infty f(x) \d x$ both converge, or both diverge. It is impossible for one of them to diverge, and the other to converge.

Notice we’ve also included the observation that a finite number of terms cannot affect the convergence or divergence of a series. In particular, our function $f$ can be different from our sequence for a finite number of terms.

Does the harmonic series $\sum _{k=1}^\infty \frac {1}{k} = 1 + \frac {1}{2} + \frac {1}{3}+ \dots$ converge or diverge?

converge diverge

By the integral test, $\sum _{k=1}^\infty \frac {1}{k}$ converges if and only if $\int _1^\infty \frac {1}{x} \d x$ converges

$\begin{align*} \int_1^\infty \frac{1}{x} \d x &= \lim_{b \to \infty} \int_1^b \frac{1}{x} \d x\\ &=\lim_{b \to \infty} \ln(b) \\ &=\infty \end{align*}$ Thus the Harmonic series must diverge.

Our last example is a “theorem” in some textbooks. We think it is just an application of the integral test. However, you’ll want to be able to apply this result to other examples!

$p$ -test Explain why the sum of the “ $p$ -series” $\sum _{n=1}^\infty \frac {1}{n^p}$ converges if and only if $p > 1.$

First, recall that we have already dealt with the case that $p = 1$ , since this case is exactly the harmonic series. We have used the integral test to show that the harmonic series diverges. Now, assume that $p \ne 1$ . By the integral test, $\sum _{n=1}^\infty \frac {1}{n^p}$ converges if and only if $\int _1^\infty \answer [given]{\frac {1}{x^p}} \d x$ converges. Let’s check this, write with me: $\begin{align*} \int_1^\infty \frac{1}{x^p} \d x &= \lim_{b\to\infty} \int_1^b x^{-p} \d x\\ &= \lim_{b\to\infty} \eval{\answer[given]{\frac{x^{1-p}}{1-p}}}_1^b\\ &= \lim_{b\to\infty} \frac{b^{1-p}}{1-p} - \frac{1^{1-p}}{1-p}\\ &= \lim_{b\to\infty} \frac{b^{1-p}}{1-p} - \frac{1}{1-p} \end{align*}$ and this only converges when $p>\answer [given]{1}$ .

Estimating Series Using Improper Integrals

Another cool application of this graphical viewpoint on series involves estimating the sum of a series.

We cannot yet compute $\sum _{k=1}^\infty \frac {1}{k^2}$ exactly, and we will not learn how to do so in this course.

Say we wanted to approximate this sum within an error of $\frac {1}{100}$ . How many terms should we sum? Is it enough to sum the first ten terms? The first hundred? We want to find a number $N$ where we can be sure that $\sum _{k=1}^\infty \frac {1}{k^2}-\sum _{k=1}^N \frac {1}{k^2} < \frac {1}{100}.$ Again, we’re looking for a number $N$ so that the difference between the actual (but unknown) sum of the series and the $N$ -th partial sum is less than $\frac {1}{100}$ . Now, $\sum _{k=1}^\infty \frac {1}{k^2}- \sum _{k=1}^N \frac {1}{k^2} = \sum _{N+1}^\infty \frac {1}{k^2}.$ Set $R_N$ to be $R_N = \sum _{N+1}^\infty \frac {1}{k^2}.$ This $R_N$ is often called the “remainder” or “tail” of the series, because it is what is left over after we remove the $N$ -th partial sum. Consider the following graph, where the rectangles correspond to the terms of $R_N$ .

Shifting these rectangles over one unit, we have the following.

Now we know $R_N < \int _N^\infty \frac {1}{x^2} \d x .$ In other words, the remainder of the series must be less than the given integral. So, if we can find a whole number $N$ for which $\int _N^\infty \frac {1}{x^2} \d x < \frac {1}{100},$ we will be sure that we have summed enough terms in the series to get to within $\frac {1}{100}$ .

What is the least natural number $N$ for $\sum _{k=1}^N \frac {1}{k^2}$ so that $\sum _{k=1}^\infty \frac {1}{k^2} - \sum _{k=1}^N \frac {1}{k^2} < \frac {1}{100} ?$

From our work above, we see that we need to find $N$ such that $\int _N^\infty \frac {1}{x^2} \d x < \answer [given]{\frac {1}{100}}.$ Write with me. $\begin{align*} \int_N^\infty \frac{1}{x^2} \d x &< \frac{1}{100}\\ \lim_{b \to \infty} \int_N^b \frac{1}{x^2} \d x &< \frac{1}{100}\\ \lim_{b \to \infty} \eval{\answer[given]{\frac{-1}{x}}}_N^b& < \frac{1}{100}\\ \lim_{b \to \infty} \frac{1}{N} - \frac{1}{b}& < \frac{1}{100}\\ \frac{1}{N} &< \frac{1}{100}\\ N&>\answer[given]{100}. \end{align*}$ So we need $N$ to be at least $\answer [given]{101}$ .

This shows that if we sum $101$ terms of the series $\sum _{k=1}^\infty \frac {1}{k^2}$ , we will get within $\frac {1}{100}$ of the true answer.

Using a computer to sum the first $101$ terms we get $\sum _{k=1}^{101} \frac {1}{k^2} \approx 1.6350.$ Using advanced concepts like Fourier Analysis, it turns out this series actually converges to $\frac {\pi ^2}{6}$ , and we know that $\frac {\pi ^2}{6} \approx 1.6449$ . The difference between our calculated value and the actual value is just barely less than $\frac {1}{100}$ .