The integral test

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \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]{\,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}}}\hspace {0in}} \newcommand {\point }[1]{\left (#1\right )} \newcommand {\pt }[1]{\mathbf {#1}} \newcommand {\Lim }[2]{\lim _{\point {#1} \to \point {#2}}} \DeclareMathOperator {\proj }{\mathbf {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 }$

Certain infinite series can be studied using improper integrals.

In order to study the convergence of a series $\sum _{k=k_0}^{\infty } a_k$ , our first attempt to determine whether the series converges is to form the sequence of partial sums $\{s_n\}_{n=k_0}$ since we know that the series $\sum _{k=k_0}^{\infty } a_k$ converges if and only if $\lim _{n \to \infty } s_n$ exists. In the case of geometric or telescoping series, we were able to find an explicit formula for $s_n$ , and analyze $\lim _{n \to \infty } s_n$ by explicit computation. However, we cannot always find such an explicit formula, and when this is the case, we try to use properties of the terms in the sequence $\{a_n\}$ to determine whether $\lim _{n \to \infty } s_n$ exists. Our first result was the divergence test, which states

If $\lim _{n \to \infty } a_n \neq 0$ , then $\sum _{k=k_0} a_k$ diverges.

However, there are still some divergent series that the divergence test does not pick out! We begin this section with such an example that shows how there is a connection between certain special types of series and improper integrals.

Consider the series $\sum _{k=1}^{\infty } \frac {1}{k}$ . This series is called the harmonic series and will be important in the coming material. Notice that this series isis not geometric. It isis not telescoping, and $\lim _{n \to \infty } \frac {1}{n} = \answer {0}$ , so the divergence test isis not conclusive. So what should we do?

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

As it turns out, there is a nice way to visualize the sum $\sum _{k=1}^\infty \frac {1}{k}$ too! One such way is to to make rectangles whose areas are equal to the terms in the sequence.

Note that the height of the $k$ -th rectangle is precisely $\frac {1}{k}$ and the width of all of the rectangles is $1$ , so the area of the $k$ -th rectangle is $\frac {1}{k}$ .

Now, in order to conclude whether $\sum _{k=1}^{\infty } a_k$ converges, we must analyze $\lim _{n \to \infty } s_n$ . Note that $s_n$ has a nice visual interpretation as the sum of the areas of the first $n$ rectangles, but since we don’t have an explicit formula for $s_n$ we can try to establish that

$\{s_n\}$ is bounded and monotonic, and hence $\lim _{n \to \infty } s_n$ exists.
$\{s_n\}$ is unbounded so $\lim _{n \to \infty } s_n$ does not exist.

How can we establish this? The previous image might 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. For instance, if we add a plot of $1/x$ to our picture above

we see that $\int _1^n \frac {1}{x} \d x < \sum _{k=1}^n \frac {1}{k}.$

Notice that the sum on the righthand side is simply $s_n$ . Since $\int _1^{\infty } \frac {1}{x} \d x$ is an improper integral, so we need to determine whether $\lim _{n \to \infty } \left [ \int _1^n \frac {1}{x} \d x \right ]$ exists. Notice

$\lim _{n \to \infty } \left [ \int _1^n \frac {1}{x} \d x \right ] = \lim _{n \to \infty } \eval {\ln (x)}_1^n = \infty .$

This means that $\{s_n\}$ is not bounded and hence $\lim _{n \to \infty } s_n = \infty$ and $\sum _{k=1}^{\infty } \frac {1}{k}$ must diverge.

To argue why $\{s_n\}$ is not bounded a little more formally, note that if $\{s_n\}$ were bounded, then all of the terms in the sequence $\{s_n\}$ would be less than some number $M$ . Since $\lim _{n \to \infty } \left [ \int _1^n \frac {1}{x} \d x \right ] = \infty$ , this means that, for some value of $N$ after which all values of $n>N$ , $\int _1^n \frac {1}{x} \d x > M$ , and since $\int _1^n \frac {1}{x} \d x < s_n$ for all $n$ , we may conclude that $s_n >M$ for all $n\geq N$ and hence $\{s_n\}$ is unbounded.

Now, let’s take a step back and see what we really needed in the previous example.

We needed to find a function for which the area under the curve over any particular interval $[n,n+1]$ was less than the area of the rectangle whose height is $a_k$ to establish a lower bound for each $s_n$ . Note that we can always do this if $f(x)$ is eventually positive and since we may view each $a_k$ as the area of the rectangle that coincides with $f(x)$ at its lefthandrighthand endpoint.
We needed the function to be “eventually continuous” so the improper integral $\int _{a}^{\infty } f(x) \d x$ can be computed as the limit of a single definite integral.

By “eventually” above, we really mean that $f(x)$ should be continuous, positive, and decreasing on some interval $[a,\infty )$ for some $a>0$ ; it doesn’t need to happen right away, but it should hold for all real large enough $x$ -values. This leads us to an interesting observation.

Let $f(x)$ be an eventually continuous, positive, and decreasing function 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 even better.

Consider the sequence $\{a_n\}_{n=1}$ where $a_n = \frac {1}{n^2}$ . We can write out a few terms

$1, \, \frac {1}{4}, \, \frac {1}{9}, \, \frac {1}{16}, \, \frac {1}{25}, \,\dots$ and try the same idea of visualizing the series $\sum _{k=1}^{\infty } \frac {1}{k^2}$ as an area.

The shaded area above exactly represents to the sum $\sum _{k=1}^\infty \frac {1}{k^2}.$ We again visually compare the sum of an infinite series to the value of an improper integral by adding a plot of $1/x^2$ to our picture above.

Note that we have a slight annoyance if we consider $\int _{0}^{\infty } \frac {1}{x^2} \d x$ since $\frac {1}{x^2}$ has a vertical asymptote at $x=0$ . This is easily avoided if we instead consider $\frac {1}{x^2}$ on the interval $[1,\infty )$ . This requires that we consider the rectangles on that interval too. We update our picture.

Notice that the sum of the areas of the rectangles now is the series $\sum _{k=0}^{\infty } \frac {1}{k^2}$ $\sum _{k=1}^{\infty } \frac {1}{k^2}$ $\sum _{k=2}^{\infty } \frac {1}{k^2}$ .

This is not an issue, because we know

The value of the lower index of summation does not affect whether an infinite series converges or diverges.

That is, if we can show $\sum _{k=2}^{\infty } \frac {1}{k^2}$ converges, then $\sum _{k=1}^{\infty } \frac {1}{k^2}$ must also converge. Now to determine whether $\sum _{k=2}^{\infty } \frac {1}{k^2}$ converges, let $s_n = \sum _{k=2}^n \frac {1}{k^2}$ . We must determine if $\lim _{n \to \infty } s_n$ exists. Since we expect $\int _1^{\infty } f(x) \d x$ to converge, we expect $\sum _{k=2}^{\infty } \frac {1}{k^2}$ to converge. To show this, we should establish $\{s_n\}$ is bounded and monotonic, and hence $\lim _{n \to \infty } s_n$ exists $\{s_n\}$ is unbounded so $\lim _{n \to \infty } s_n$ does not exist .

Notice that since $\frac {1}{k^2} > 0$ for all $k$ , and $s_n = \sum _{k=1}^n$ , the sequence $\{s_n\}$ is increasingdecreasing .

To show that it is also bounded, note that for any $n \geq 2$ , we can observe from the picture that

$s_n = \sum _{k=1}^n\frac {1}{k^2} \leq \int _1^n \frac {1}{x^2} \d x.$ In particular, since we shade more area if we increase $n$ , we have that $\int _1^n \frac {1}{x^2} \d x \leq \int _1^\infty \frac {1}{x^2} \d x$ . By a routine calculation of the latter improper integral, we can show

$\int _1^\infty \frac {1}{x^2} \d x = 1,$

and we thus have

$s_n = \sum _{k=1}^n\frac {1}{k^2} \leq \int _1^n \frac {1}{x^2} \d x \leq 1 \textrm { for all } n \geq 2.$

Hence $\{s_n\}$ is both bounded and monotonic, so $\lim _{n \to \infty } s_n$ exists, and $\sum _{k=2}^{\infty } \frac {1}{k^2}$ converges. Since $\sum _{k=2}^{\infty } \frac {1}{k^2}$ converges, $\sum _{k=1}^{\infty } \frac {1}{k^2}$ will also converge since the index where we start the summation does not affect whether the series converges.

Note that from the above, the value of the improper integral the value of the series; indeed, by visualizing the series as the sum of the areas of the rectangles in the image, we see that the value of the series $\sum _{k=2}^{\infty } \frac {1}{k^2}$ should be the value of the improper integral $\int _2^{\infty } \frac {1}{x^2}$ .

Now, let’s take a step back and see what we really needed in the this example.

We needed to find a function for which the area under the curve over any particular interval $[n,n+1]$ was greater than the area of the rectangle whose height is $a_k$ to establish a lower bound for each $s_n$ . Note that we can always do this if $f(x)$ is eventually positive and since we may view each $a_k$ as the area of the rectangle that coincides with $f(x)$ at its lefthandrighthand endpoint.
We needed to establish that the sequence of partial sums is eventually increasing. This must happen if all of the $a_k$ are positivenegative .
We needed the function to be “eventually continuous” so the improper integral $\int _{a}^{\infty } f(x) \d x$ can be computed as the limit of a single definite integral.

By “eventually” above, we really mean that $f(x)$ should be continuous, positive, and decreasing on some interval $[a,\infty )$ ; it doesn’t need to happen right away, but it should hold for all real large enough $x$ -values. This leads us to an interesting observation.

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

The Integral Test

The observations from the previous examples give us a new convergence test called the integral test:

Integral Test Suppose that $\{a_n\}_{n=n_0}$ is a sequence, and suppose that $f(x)$ is an eventually continuous, positive, and decreasing function with $f(n)=a_n$ for all $n \geq N$ , where $N$ is an integer. Then, $\sum _{k=n_0}^\infty a_k\text { and }\int _N^\infty f(x) \d x$ either both converge or both diverge.

Determine whether $\sum _{k=1}^{\infty } \frac {k}{(2k^2+1)^2}$ converges.

Note that $\lim _{n \to \infty } \frac {n}{(2n+1)^2} =0$ so the divergence test is inconclusive. Also, this is not a geometric series. Let’s try the integral test with $f(x) = \frac {x}{(2x+1)^2}$ . Notice that the function $f(x) = \frac {x}{(2x^2+1)^2}$ is continuous and positive on $[1,\infty )$ . We can check where it is decreasing part by computing $f'(x) = \frac {-6x^2+1}{(2x^2+1)^3} < 0$ . So, $f'(x)$ is certainly negative for $x >1$ and hence $f(x)$ is also decreasing on $[1,\infty )$ .

Notice that, after performing a substitution if necessary,

$\int \frac {x}{(2x+1)^2} \d x = -\frac {1}{8x^2+4} +C$

so $\lim _{b \to \infty } \eval { -\frac {1}{8x^2+4}}_1^b = \frac {1}{8}$ and hence the improper integral $\int _1^{\infty } \frac {x}{(2x+1)^2} \d x$ . Thus, $\sum _{k=1}^{\infty } \frac {k}{(2k^2+1)^2}$ convergesdiverges .

The next examples synthesizes some concepts we have seen thus far.

Let $a_n = \frac {1}{n^2+4}$ for $n \geq 1$ and let $s_n = \sum _{k=1}^n a_k$ . Determine if $\{a_n\}_{n=1}$ and $\{s_n\}_{n=1}$ are bounded and/or monotonic.

Note that $\{a_n\}_{n=1}$ is decreasing (and hence monotonic) because denominator increases as $n$ increases. Hence, the fraction becomes smaller as $n$ grows larger.

$\{a_n\}_{n=1}$ is also bounded since $0 \leq a_n \leq 1$ for all $n$ (i.e. no terms are larger than $1$ or smaller than $0$ ).

For $s_n$ , note that

$s_n = s_{n-1} +a_n = s_{n-1} + \frac {1}{n^2+4}.$

Since $\frac {1}{n^2+4} >0$ , we have that $s_{n+1} > s_n$ for all $n$ , so $\{s_n\}_{n=1}$ is increasing (and hence monotonic). To determine if $\{s_n\}_{n=1}$ bounded, we can determine whether $\lim _{n \to \infty } s_n$ exists. Indeed, since $\{s_n\}_{n=1}$ is increasing, we have that $s_n < \lim _{n \to \infty } s_n$ , so if the limit exists, it serves as an upper bound for all of the terms in the sequence.

We can write out the limit in this case and find

$\lim _{n \to \infty } s_n = \lim _{n \to \infty } \left [ \sum _{k=1}^n \frac {1}{k^2+4} \right ] = \sum _{k=1}^{\infty } \frac {1}{k^2+4}$

Thus, all we have to do is determine if the infinite series above converges.

To do this, we can apply the integral test. Let $f(x)= \frac {1}{x^2+4}$ . Notice that $f(x)$ is positive, continuous, and decreasing on $[1,\infty )$ , and

$\lim _{b \to \infty } \left [ \int _0^b \frac {1}{x^2+4} \right ]= \lim _{b \to \infty } \eval {\frac {1}{2} \arctan \left (\frac {x}{2}\right )}_1^b =\frac {\pi }{4}- \frac {1}{2}\arctan \left (\frac {1}{2}\right ).$

Since the improper integral converges, the series $\sum _{k=1}^{\infty } \frac {1}{k^2+4}$ converges. Hence, $\lim _{n \to \infty }s_n$ exists so $\{s_n\}_{n=1}$ is bounded.

p-Series

A very important type of series for future sections is $\sum _{k=1}^\infty \frac {1}{k^p}$ , where $p>0$ . We call a series that can be brought into this form a $p$ -series. We want to determine for which values of $p$ these series converge and diverge.

Notice that in our model examples, both series were $p$ -series.

The harmonic series $\sum _{k=1}^{\infty } \frac {1}{k}$ is a $p$ -series with $p = \answer {1}$ . It convergesdiverges .
The series $\sum _{k=1}^{\infty } \frac {1}{k^2}$ is a $p$ -series with $p = \answer {2}$ . It convergesdiverges .

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, $\sum _{k=1}^\infty \frac {1}{k^p}$ converges if and only if $\int _1^\infty \answer [given]{\frac {1}{x^p}} \d x$ converges. Let’s check this, write with me: $\begin{align*} \int_1^\infty \frac{1}{x^p} \d x &= \lim_{b\to\infty} \int_1^b x^{-p} \d x\\ &= \lim_{b\to\infty} \eval{\answer[given]{\frac{x^{1-p}}{1-p}}}_1^b\\ &= \lim_{b\to\infty} \left[ \frac{b^{1-p}}{1-p} - \frac{1^{1-p}}{1-p} \right] \\ &= \lim_{b\to\infty} \left[\frac{b^{1-p}}{1-p}\right] - \frac{1}{1-p} \end{align*}$ Note that if $p >1$ , the exponent $1-p$ is negative so $\lim _{b\to \infty } \left [\frac {b^{1-p}}{1-p}\right ] = \answer {0}$ . If $p <1$ , the exponent $1-p$ is positive so $\lim _{b\to \infty } \left [\frac {b^{1-p}}{1-p}\right ] = \answer {\infty }$ . Thus, $\int _1^\infty \frac {1}{x^p} \d x$ only converges when $p>1$ and thus $\sum _{k=1}^\infty \frac {1}{k^p}$ converges if and only if $p>1$ .

This result is important enough to list as a theorem.

$p$ -series test The series $\sum _{k=1}^\infty \frac {1}{k^p}$ converges if and only if $p > 1$ .

Which of the following series converge? (Select all that apply)

$\sum _{k=2}^{\infty } \frac {1}{\sqrt {k}}$ $\sum _{k=3}^{\infty } \frac {1}{k^{3/2}}$ $\sum _{k=5}^{\infty } k^{-4}$ $\sum _{k=1}^{\infty } \frac {1}{k^{.1}}$