8.3 The Divergence and Integral Tests
Two first tests for a series $\sum a_n$: if the terms do not shrink to $0$ the series must diverge, and if the terms come from a positive decreasing function the series behaves exactly like an improper integral.
In Section 8.2 we could add up geometric and telescoping series exactly, because their partial sums $S_N$ had a closed form. Those series are the exception. For a series as innocent-looking as $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}=1+\frac14+\frac19+\frac1{16}+\cdots$ there is no tidy formula for $S_N$, and a computer only gives us finitely many partial sums: $S_{1000}\approx1.643935$ and $S_{1\,000\,000}\approx1.644933$. Do these numbers settle down, or are they creeping upwards for ever? A computer cannot tell us — the harmonic series $\sum 1/n$ also creeps very slowly ($S_{1\,000\,000}\approx14.39$), yet it diverges.
So we need convergence tests: arguments that decide whether $\sum a_n$ converges by looking at the terms $a_n$, without ever finding the sum. This section gives the first two. The divergence test is a quick filter that can only ever prove divergence. The integral test compares the series with an improper integral (Section 5.5), settles the whole family of $p$-series $\sum 1/n^p$ at once, and — as a bonus — tells us how accurately a partial sum approximates the true sum.
The Divergence Test
If $\sum a_n$ converges to $S$, then the partial sums satisfy $S_N\to S$, and so does $S_{N-1}$. Since each term is the difference of two consecutive partial sums, $a_N=S_N-S_{N-1}\to S-S=0$. Read the other way round, this is a test.
📐 Theorem 8.11 — The divergence test ($n$th-term test)
If $\displaystyle\sum_{n=1}^{\infty}a_n$ converges, then $\displaystyle\lim_{n\to\infty}a_n=0$. Equivalently: if $\displaystyle\lim_{n\to\infty}a_n\ne0$, or the limit does not exist, then $\displaystyle\sum_{n=1}^{\infty}a_n$ diverges.
✏️ Example 1 — Using the divergence test
For each series, decide what the divergence test tells us.
(a) $\displaystyle\sum_{n=1}^{\infty}\frac{n^2}{3n^2+1}$ (b) $\displaystyle\sum_{n=1}^{\infty}\cos\frac1n$ (c) $\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{n}{n+1}$ (d) $\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt n}$
💡 Show solution▶
(a) Divide top and bottom by $n^2$: $\displaystyle\lim_{n\to\infty}\frac{n^2}{3n^2+1}=\lim_{n\to\infty}\frac{1}{3+1/n^2}=\frac13\ne0$. The series diverges. (Adding infinitely many terms each close to $\tfrac13$ cannot give a finite total.)
(b) As $n\to\infty$, $1/n\to0$ and cosine is continuous, so $\cos(1/n)\to\cos0=1\ne0$. The series diverges.
(c) The size of the terms is $\dfrac{n}{n+1}\to1$, but the sign alternates, so the terms approach $+1$ along even $n$ and $-1$ along odd $n$. The limit does not exist — in particular it is not $0$ — so the series diverges.
(d) $\dfrac{1}{\sqrt n}\to0$. The test is inconclusive: it gives no information at all. (We will see below that this series diverges — it is a $p$-series with $p=\tfrac12$.)
⚠️ Analysis — the converse is false
1. Terms tending to $0$ is necessary, not sufficient. Consider $\displaystyle\sum_{n=1}^{\infty}\ln\left(1+\frac1n\right)$. Its terms tend to $\ln1=0$, but $\ln\left(1+\tfrac1n\right)=\ln(n+1)-\ln n$, so the series telescopes: $$S_N=(\ln2-\ln1)+(\ln3-\ln2)+\cdots+\big(\ln(N+1)-\ln N\big)=\ln(N+1)\to\infty.$$ The terms shrink, yet the sum grows without bound. The harmonic series $\sum 1/n$ behaves the same way (Theorem 8.13 below).
2. The test can only say “diverges”. If $a_n\to0$, write “the divergence test is inconclusive” — never “so the series converges”.
3. Use it first. Checking $\lim a_n$ takes seconds, and it saves you from trying a harder test on a series such as $\sum\frac{n}{2n+1}$ that obviously diverges.
The Integral Test
Suppose the terms are values of a function: $a_n=f(n)$, where $f$ is positive and decreasing. Draw a rectangle of width $1$ and height $a_n$ for each term. Then the sum of the series is the total area of the rectangles, and we can compare it with the area under the curve $y=f(x)$.
- Put the rectangle for $a_n$ on $[n,n+1]$. Because $f$ is decreasing, its top is at the left-end height, so it sits above the curve: $a_n\ge\displaystyle\int_n^{n+1}f(x)\,dx$. Adding $n=1,\dots,N$: $$S_N\ \ge\ \int_1^{N+1}f(x)\,dx.$$
- Put the rectangle for $a_n$ on $[n-1,n]$ instead. Now its top is at the right-end height, so it sits below the curve: $a_n\le\displaystyle\int_{n-1}^{n}f(x)\,dx$. Adding $n=2,\dots,N$: $$S_N-a_1\ \le\ \int_1^{N}f(x)\,dx.$$
So $\displaystyle\int_1^{N+1}f(x)\,dx\le S_N\le a_1+\int_1^{N}f(x)\,dx$. If the improper integral is finite, the increasing sequence $S_N$ is bounded above and therefore converges (Monotone Convergence Theorem, Section 8.1). If the integral is infinite, $S_N$ is pushed up to $\infty$. Play with Interactive 1 to see both pictures.
📐 Theorem 8.12 — The integral test
Let $k$ be a positive integer and suppose that on $[k,\infty)$ the function $f$ is
- positive,
- continuous, and
- decreasing,
⚠️ Analysis — reading the theorem carefully
1. “Eventually” is enough. Changing or deleting finitely many terms never changes whether a series converges (only its sum). So we only need $f$ to be positive, continuous and decreasing on some interval $[k,\infty)$ — we say $f$ is eventually decreasing. To check “decreasing”, show $f'(x)\lt 0$ for $x\ge k$.
2. The sum is not the integral. For $\sum_{n=1}^\infty 1/n^2$ we have $\int_1^\infty x^{-2}dx=1$, but the sum is about $1.645$. The test tells us only whether the series converges; the inequalities above (and the remainder estimate later) tell us roughly what the sum is.
3. All three hypotheses matter. For $\sum\frac{2+\sin n}{n^2}$ the natural $f(x)=\frac{2+\sin x}{x^2}$ is positive and continuous but keeps wiggling up and down, so it is not eventually decreasing and the integral test does not apply (the comparison test of Section 8.4 handles it in one line). For series with negative terms the test does not apply at all.
🧭 Using the integral test
- Check the divergence test first. If $a_n\not\to0$ you are done.
- Write $a_n=f(n)$ and choose $k$. Check that $f$ is positive and continuous on $[k,\infty)$ and that $f'(x)\lt 0$ there (or that $f$ is decreasing for an obvious reason, such as an increasing positive denominator).
- Evaluate $\displaystyle\int_k^\infty f(x)\,dx=\lim_{t\to\infty}\int_k^t f(x)\,dx$.
- Conclude: finite integral $\Rightarrow$ series converges; infinite integral $\Rightarrow$ series diverges.
✏️ Example 2 — A divergent series whose terms tend to zero
Determine whether $\displaystyle\sum_{n=1}^{\infty}\frac{n}{n^2+1}$ converges or diverges.
💡 Show solution▶
Divergence test $\dfrac{n}{n^2+1}=\dfrac{1/n}{1+1/n^2}\to0$, so this test is inconclusive. We need something stronger.
Hypotheses Let $f(x)=\dfrac{x}{x^2+1}$. It is positive and continuous on $[1,\infty)$, and $$f'(x)=\frac{(x^2+1)-x(2x)}{(x^2+1)^2}=\frac{1-x^2}{(x^2+1)^2}\le0\quad\text{for }x\ge1,$$ so $f$ is decreasing on $[1,\infty)$.
Integral With $u=x^2+1$, $du=2x\,dx$: $$\int_1^{\infty}\frac{x}{x^2+1}\,dx=\lim_{t\to\infty}\Big[\tfrac12\ln(x^2+1)\Big]_1^t=\lim_{t\to\infty}\left(\tfrac12\ln(t^2+1)-\tfrac12\ln2\right)=\infty.$$
Conclusion The integral diverges, so by the integral test $\displaystyle\sum_{n=1}^{\infty}\frac{n}{n^2+1}$ diverges. For large $n$ the terms behave like $\frac1n$, and the harmonic series diverges too — an idea we exploit fully in Section 8.4.
✏️ Example 3 — A logarithm in the denominator
Show that $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n(\ln n)^2}$ converges, but $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln n}$ diverges.
💡 Show solution▶
Hypotheses On $[2,\infty)$ both $f(x)=\dfrac{1}{x(\ln x)^2}$ and $g(x)=\dfrac{1}{x\ln x}$ are continuous and positive (since $\ln x\ge\ln2\gt 0$), and both are decreasing because their denominators are products of positive increasing factors.
First integral Substitute $u=\ln x$, $du=\dfrac{dx}{x}$: $$\int_2^{\infty}\frac{dx}{x(\ln x)^2}=\lim_{t\to\infty}\left[-\frac{1}{\ln x}\right]_2^t=\lim_{t\to\infty}\left(\frac{1}{\ln2}-\frac{1}{\ln t}\right)=\frac{1}{\ln2}\approx1.4427.$$ Finite, so $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n(\ln n)^2}$ converges.
Second integral The same substitution gives $$\int_2^{\infty}\frac{dx}{x\ln x}=\lim_{t\to\infty}\Big[\ln(\ln x)\Big]_2^t=\lim_{t\to\infty}\big(\ln(\ln t)-\ln(\ln2)\big)=\infty,$$ so $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln n}$ diverges.
Comment The two series differ only by one extra factor of $\ln n$ — a function that grows more slowly than any power of $n$ — yet that is enough to tip the balance. Convergence can be a delicate matter.
✏️ Example 4 — When $f$ is only eventually decreasing
Determine whether $\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n^2}$ converges.
💡 Show solution▶
Where is $f$ decreasing? Let $f(x)=\dfrac{\ln x}{x^2}$. By the quotient rule $$f'(x)=\frac{x^2\cdot\frac1x-2x\ln x}{x^4}=\frac{1-2\ln x}{x^3},$$ which is negative when $\ln x\gt\tfrac12$, i.e. $x\gt\sqrt e\approx1.65$. So $f$ increases on $[1,\sqrt e\,]$ and is decreasing on $[2,\infty)$. It is also continuous and positive on $[2,\infty)$, so we apply the test with $k=2$.
Integral By parts with $u=\ln x$, $dv=x^{-2}dx$ ($du=\tfrac1x dx$, $v=-\tfrac1x$): $$\int_2^{t}\frac{\ln x}{x^2}\,dx=\left[-\frac{\ln x}{x}\right]_2^t+\int_2^t\frac{dx}{x^2}=-\frac{\ln t}{t}+\frac{\ln2}{2}-\frac1t+\frac12.$$ Since $\dfrac{\ln t}{t}\to0$ (L'Hôpital), the improper integral equals $\dfrac{1+\ln2}{2}\approx0.8466$.
Conclusion $\displaystyle\sum_{n=2}^{\infty}\frac{\ln n}{n^2}$ converges; adding the single term $n=1$ (which is $0$) does not change that. So $\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n^2}$ converges.
Check Numerically the sum is about $0.9375$, and the bounds $\int_2^{N+1}f\le S_N-a_1\le a_2+\int_2^N f$ with $N\to\infty$ give $0.8466\le S\le 0.1733+0.8466=1.0199$. ✓
The $p$-Series
📘 Definition 8.5 — $p$-series
For a real number $p$, the series $$\sum_{n=1}^{\infty}\frac{1}{n^p}=1+\frac{1}{2^p}+\frac{1}{3^p}+\frac{1}{4^p}+\cdots$$ is called a $p$-series. The case $p=1$ is the harmonic series.
📐 Theorem 8.13 — Convergence of $p$-series
The $p$-series $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^p}$ converges if $p\gt 1$ and diverges if $p\le1$.
💡 Show proof▶
$p\le0$ Then $\dfrac1{n^p}=n^{-p}$ with $-p\ge0$, so the terms equal $1$ (if $p=0$) or tend to $\infty$. They do not tend to $0$, so the series diverges by the divergence test.
$p\gt 0$ $f(x)=x^{-p}$ is positive and continuous on $[1,\infty)$, and $f'(x)=-px^{-p-1}\lt 0$, so the integral test applies. For $p\ne1$, $$\int_1^{\infty}x^{-p}\,dx=\lim_{t\to\infty}\frac{t^{1-p}-1}{1-p}=\begin{cases}\dfrac{1}{p-1}, & p\gt 1\ \ (\text{since }t^{1-p}\to0),\\[2mm] \infty, & 0\lt p\lt 1\ \ (\text{since }t^{1-p}\to\infty).\end{cases}$$ For $p=1$, $\int_1^\infty\frac{dx}{x}=\lim_{t\to\infty}\ln t=\infty$.
Conclusion The integral (and hence the series) converges exactly when $p\gt 1$. ∎
✏️ Example 5 — Recognising $p$-series
Decide whether each series converges. (a) $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n\sqrt n}$ (b) $\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt[3]{n^2}}$ (c) $\displaystyle\sum_{n=1}^{\infty}n^{-1.001}$ (d) $\displaystyle\sum_{n=1}^{\infty}\frac{5}{n^{0.9}}$
💡 Show solution▶
(a) $n\sqrt n=n^{3/2}$, so this is a $p$-series with $p=\tfrac32\gt 1$: converges.
(b) $\sqrt[3]{n^2}=n^{2/3}$, so $p=\tfrac23\le1$: diverges.
(c) $p=1.001\gt 1$: converges — although extremely slowly. Its sum is about $1000.6$, and you would need astronomically many terms to see it settle.
(d) $\displaystyle\sum\frac{5}{n^{0.9}}=5\sum\frac1{n^{0.9}}$, and a nonzero constant multiple of a divergent series diverges. Here $p=0.9\le1$: diverges.
Estimating the Sum: the Remainder
Knowing that $\sum 1/n^2$ converges is good; knowing its value to three decimal places is better. If we stop after $N$ terms, the error we make is everything we left out.
📘 Definition 8.6 — Remainder of a series
If $\displaystyle\sum_{n=1}^{\infty}a_n$ converges to $S$, its remainder after $N$ terms is $$R_N=S-S_N=a_{N+1}+a_{N+2}+a_{N+3}+\cdots=\sum_{n=N+1}^{\infty}a_n,$$ the error made when the partial sum $S_N$ is used to approximate $S$.
The rectangle argument, applied to the tail from $N+1$ onwards, traps $R_N$ between two integrals.
📐 Theorem 8.14 — Remainder estimate for the integral test
Suppose $f$ is positive, continuous and decreasing on $[N,\infty)$, that $a_n=f(n)$ for $n\ge N$, and that $\sum a_n$ converges. Then
$$\int_{N+1}^{\infty}f(x)\,dx\ \le\ R_N\ \le\ \int_{N}^{\infty}f(x)\,dx,$$
and therefore
$$S_N+\int_{N+1}^{\infty}f(x)\,dx\ \le\ S\ \le\ S_N+\int_{N}^{\infty}f(x)\,dx.$$
✏️ Example 6 — Approximating $\sum 1/n^2$
(a) Use $S_{10}$ and Theorem 8.14 to find bounds for $S=\displaystyle\sum_{n=1}^{\infty}\frac1{n^2}$. (b) Estimate $S$ by the midpoint of the bounds, with an error bound, and compare with the exact value $\dfrac{\pi^2}{6}$. (c) How many terms guarantee $R_N\le0.001$?
💡 Show solution▶
Integrals $f(x)=x^{-2}$ is positive, continuous and decreasing on $[1,\infty)$, and $\displaystyle\int_N^{\infty}\frac{dx}{x^2}=\lim_{t\to\infty}\left(\frac1N-\frac1t\right)=\frac1N$.
(a) $S_{10}=1+\frac14+\frac19+\cdots+\frac1{100}\approx1.549768$. Theorem 8.14 with $N=10$: $$S_{10}+\frac1{11}\le S\le S_{10}+\frac1{10}\quad\Longrightarrow\quad 1.640677\le S\le1.649768.$$
(b) The midpoint is $1.645222$, and $S$ is within half the width of the interval, $\tfrac12\left(\tfrac1{10}-\tfrac1{11}\right)\approx0.004545$, of it. The exact value is $\dfrac{\pi^2}{6}\approx1.644934$ — the midpoint is actually off by only $0.000288$. To get that close with a raw partial sum alone you would need about $3500$ terms, because $R_N\approx\frac1N$.
(c) We need $R_N\le\displaystyle\int_N^\infty\frac{dx}{x^2}=\frac1N\le0.001$, i.e. $N\ge1000$. $N=1000$ terms suffice. (If we use the midpoint of the bounds instead, the error is at most $\tfrac12\left(\tfrac1N-\tfrac1{N+1}\right)=\tfrac{1}{2N(N+1)}$, which is $\le0.001$ already for $N=22$.)
Exercises
I
For each series find $\displaystyle\lim_{n\to\infty}a_n$. State whether the divergence test proves that the series diverges, or whether the test is inconclusive.
- 1 $\displaystyle\sum_{n=1}^{\infty}\frac{2n+1}{5n-3}$
- 2 $\displaystyle\sum_{n=1}^{\infty}\frac{n}{\sqrt{n^2+4}}$
- 3 $\displaystyle\sum_{n=1}^{\infty}\frac{n}{n^2+1}$
- 4 $\displaystyle\sum_{n=1}^{\infty}(-1)^n$
- 5 $\displaystyle\sum_{n=1}^{\infty}n\sin\frac1n$
- 6 $\displaystyle\sum_{n=1}^{\infty}e^{1/n}$
- 7 $\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n}$
- 8 $\displaystyle\sum_{n=1}^{\infty}\left(1-\frac2n\right)^n$
- 9 $\displaystyle\sum_{n=1}^{\infty}\tan^{-1}n$
- 10 $\displaystyle\sum_{n=1}^{\infty}\frac{n^2}{2^n}$
💡 Show solutions▶
1 Divide by $n$: $\dfrac{2+1/n}{5-3/n}\to\dfrac25\ne0$. Diverges.
2 $\dfrac{n}{\sqrt{n^2+4}}=\dfrac{1}{\sqrt{1+4/n^2}}\to1\ne0$. Diverges.
3 $\dfrac{n}{n^2+1}\to0$. Inconclusive. (Example 2 shows it diverges by the integral test.)
4 The terms alternate $-1,1,-1,\dots$; the limit does not exist. Diverges. (The partial sums are $-1,0,-1,0,\dots$.)
5 Put $h=1/n\to0^+$: $n\sin\frac1n=\dfrac{\sin h}{h}\to1\ne0$. Diverges.
6 $e^{1/n}\to e^0=1\ne0$. Diverges.
7 By L'Hôpital, $\displaystyle\lim_{x\to\infty}\frac{\ln x}{x}=\lim_{x\to\infty}\frac{1/x}{1}=0$. Inconclusive. (In fact $\frac{\ln n}{n}\ge\frac1n$ for $n\ge3$, and the series diverges — see Interactive 1.)
8 $\ln\left(1-\frac2n\right)^n=n\ln\left(1-\frac2n\right)=\dfrac{\ln(1-2h)}{h}$ with $h=\frac1n$, which tends to $-2$ by L'Hôpital. So $a_n\to e^{-2}\approx0.1353\ne0$. Diverges.
9 $\tan^{-1}n\to\dfrac{\pi}{2}\ne0$. Diverges.
10 Applying L'Hôpital twice, $\displaystyle\lim_{x\to\infty}\frac{x^2}{2^x}=\lim_{x\to\infty}\frac{2}{(\ln2)^2\,2^x}=0$. Inconclusive. (The ratio test of Section 8.6 will show that it converges, to $6$.)
II
Check that the integral test applies (state $f$, the interval $[k,\infty)$, and why $f$ is positive, continuous and decreasing there), then use it to decide whether the series converges or diverges.
- 1 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{2n+3}$
- 2 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{(n+1)^3}$
- 3 $\displaystyle\sum_{n=1}^{\infty}ne^{-n^2}$
- 4 $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln n}$ (redo it without looking at Example 3)
- 5 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2+4}$
- 6 $\displaystyle\sum_{n=2}^{\infty}\frac{\ln n}{n^3}$
- 7 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n+5}}$
- 8 $\displaystyle\sum_{n=1}^{\infty}\frac{e^{1/n}}{n^2}$
- 9 $\displaystyle\sum_{n=1}^{\infty}\frac{\tan^{-1}n}{1+n^2}$
- 10 Conceptual. Explain why the integral test cannot be applied to $\displaystyle\sum_{n=1}^{\infty}\frac{2+\sin n}{n^2}$.
Check: partial sums of the series in Exercise II — do they settle or keep creeping?
💡 Show solutions▶
1 $f(x)=\dfrac1{2x+3}$ is positive and continuous on $[1,\infty)$ and decreasing (the denominator increases). $\displaystyle\int_1^\infty\frac{dx}{2x+3}=\lim_{t\to\infty}\tfrac12\big(\ln(2t+3)-\ln5\big)=\infty$. Diverges.
2 $f(x)=(x+1)^{-3}$: positive, continuous, decreasing on $[1,\infty)$. $\displaystyle\int_1^\infty(x+1)^{-3}dx=\lim_{t\to\infty}\left[-\frac{1}{2(x+1)^2}\right]_1^t=\frac18$. Converges.
3 $f(x)=xe^{-x^2}\gt 0$, continuous, and $f'(x)=e^{-x^2}(1-2x^2)\lt 0$ for $x\ge1$. With $u=x^2$: $\displaystyle\int_1^\infty xe^{-x^2}dx=\lim_{t\to\infty}\left[-\tfrac12e^{-x^2}\right]_1^t=\frac{1}{2e}\approx0.1839$. Converges.
4 $f(x)=\dfrac{1}{x\ln x}$ is positive, continuous and decreasing on $[2,\infty)$. With $u=\ln x$: $\displaystyle\int_2^\infty\frac{dx}{x\ln x}=\lim_{t\to\infty}\big(\ln\ln t-\ln\ln2\big)=\infty$. Diverges.
5 $f(x)=\dfrac1{x^2+4}$: positive, continuous, decreasing on $[1,\infty)$. $\displaystyle\int_1^\infty\frac{dx}{x^2+4}=\lim_{t\to\infty}\left[\tfrac12\tan^{-1}\tfrac x2\right]_1^t=\frac{\pi}{4}-\frac12\tan^{-1}\frac12\approx0.5536$. Converges.
6 $f(x)=\dfrac{\ln x}{x^3}$ is positive and continuous on $[2,\infty)$, and $f'(x)=\dfrac{1-3\ln x}{x^4}\lt 0$ for $x\gt e^{1/3}\approx1.40$, so it is decreasing on $[2,\infty)$. By parts ($u=\ln x$, $dv=x^{-3}dx$): $\displaystyle\int\frac{\ln x}{x^3}dx=-\frac{\ln x}{2x^2}-\frac{1}{4x^2}+C$, so $\displaystyle\int_2^\infty\frac{\ln x}{x^3}dx=\frac{\ln2}{8}+\frac1{16}\approx0.1491$. Converges.
7 $f(x)=(x+5)^{-1/2}$: positive, continuous, decreasing on $[1,\infty)$. $\displaystyle\int_1^\infty(x+5)^{-1/2}dx=\lim_{t\to\infty}\left(2\sqrt{t+5}-2\sqrt6\right)=\infty$. Diverges.
8 $f(x)=\dfrac{e^{1/x}}{x^2}$ is positive and continuous on $[1,\infty)$, and decreasing as a product of two positive decreasing functions $e^{1/x}$ and $x^{-2}$. With $u=\frac1x$, $du=-\frac{dx}{x^2}$: $\displaystyle\int_1^\infty\frac{e^{1/x}}{x^2}dx=\lim_{t\to\infty}\Big[-e^{1/x}\Big]_1^t=-1+e=e-1\approx1.718$. Converges.
9 $f(x)=\dfrac{\tan^{-1}x}{1+x^2}\gt 0$ and continuous on $[1,\infty)$; $f'(x)=\dfrac{1-2x\tan^{-1}x}{(1+x^2)^2}\lt 0$ for $x\ge1$ because $2x\tan^{-1}x\ge2\cdot\frac{\pi}{4}\gt 1$ there. With $u=\tan^{-1}x$: $\displaystyle\int_1^\infty\frac{\tan^{-1}x}{1+x^2}dx=\left[\tfrac12(\tan^{-1}x)^2\right]_1^\infty=\tfrac12\left(\tfrac{\pi^2}{4}-\tfrac{\pi^2}{16}\right)=\frac{3\pi^2}{32}\approx0.9253$. Converges.
10 $f(x)=\dfrac{2+\sin x}{x^2}$ is positive and continuous, but $f'(x)=\dfrac{\cos x}{x^2}-\dfrac{2(2+\sin x)}{x^3}$, and for large $x$ the first term dominates, so $f'$ keeps changing sign: $f$ is not eventually decreasing. The hypotheses fail, so the test cannot be used. (Since $0\lt a_n\le\frac{3}{n^2}$, the direct comparison test of Section 8.4 shows the series converges.)
III
$p$-series. Identify $p$ and decide whether the series converges or diverges.
- 1 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^5}$
- 2 $\displaystyle\sum_{n=1}^{\infty}n^{-0.99}$
- 3 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2\sqrt n}$
- 4 $\displaystyle\sum_{n=1}^{\infty}\frac{4}{\sqrt[3]{n}}$
- 5 $\displaystyle\sum_{n=1}^{\infty}\frac{\sqrt n}{n^2}$
- 6 $\displaystyle\sum_{n=1}^{\infty}\frac{n^{e}}{n^{\pi}}$
- 7 Challenge. For which values of $p$ does $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n(\ln n)^p}$ converge?
💡 Show solutions▶
1 $p=5\gt 1$: converges.
2 $p=0.99\le1$: diverges (just barely).
3 $n^2\sqrt n=n^{5/2}$, $p=\tfrac52\gt 1$: converges.
4 $4\sum n^{-1/3}$, $p=\tfrac13\le1$: diverges.
5 $\dfrac{n^{1/2}}{n^2}=\dfrac1{n^{3/2}}$, $p=\tfrac32\gt 1$: converges.
6 $\dfrac{n^e}{n^\pi}=\dfrac{1}{n^{\pi-e}}$ with $p=\pi-e\approx0.423\le1$: diverges.
7 For $p\gt 0$, $f(x)=\dfrac1{x(\ln x)^p}$ is positive, continuous and decreasing on $[2,\infty)$. With $u=\ln x$: $\displaystyle\int_2^\infty\frac{dx}{x(\ln x)^p}=\int_{\ln2}^\infty\frac{du}{u^p}$, which converges exactly when $p\gt 1$ (the same computation as in Theorem 8.13, with $u$ in place of $x$). For $p\le0$: when $n\ge3$ we have $\ln n\gt 1$, so $(\ln n)^p\le1$ and the terms are at least $\frac1n$; the partial sums are then at least those of the harmonic series (from $n=3$), so the series diverges. It converges if and only if $p\gt 1$.
IV
Remainder estimates and applications. Use Theorem 8.14 unless told otherwise.
- 1 Find $S_5$ for $\displaystyle\sum_{n=1}^{\infty}\frac1{n^3}$ and use it to give lower and upper bounds for the sum.
- 2 How many terms of $\displaystyle\sum_{n=1}^{\infty}\frac1{n^3}$ guarantee $R_N\le0.0005$?
- 3 How many terms of $\displaystyle\sum_{n=1}^{\infty}\frac1{n^4}$ guarantee an error of at most $10^{-6}$?
- 4 Given $S_{10}\approx0.981793$ for $\displaystyle\sum_{n=1}^{\infty}\frac1{n^2+1}$, find bounds for the sum.
- 5 Show that $S_3$ approximates $\displaystyle\sum_{n=1}^{\infty}ne^{-n^2}$ with error less than $10^{-4}$, and compute $S_3$.
- 6 Applied — the leaning tower. Identical blocks of length $1$ unit can be stacked so that the top of a stack of $n$ blocks overhangs the bottom block by $\frac12H_n$, where $H_n=1+\frac12+\cdots+\frac1n$. Use the integral-test bounds $\ln(n+1)\le H_n\le1+\ln n$ to find how many blocks are certainly enough, and how many are certainly needed, for an overhang of $2$ block lengths. (The exact answer is $31$.)
- 7 Applied — aftershocks. Seismologists model the number of aftershocks on day $n$ after a large earthquake by $\dfrac{200}{n^{1.2}}$ (a form of Omori's law). Show that the predicted total number of aftershocks is finite, and estimate how many occur from day $31$ onwards. What would the model predict if the exponent were $1$ instead of $1.2$?
💡 Show solutions▶
1 $S_5=1+\frac18+\frac1{27}+\frac1{64}+\frac1{125}\approx1.185662$ and $\displaystyle\int_N^\infty x^{-3}dx=\frac{1}{2N^2}$. So $S_5+\frac{1}{72}\le S\le S_5+\frac1{50}$, i.e. $1.199551\le S\le1.205662$. (The true sum is $1.202057$.)
2 Need $\dfrac{1}{2N^2}\le0.0005$, i.e. $N^2\ge1000$, $N\ge31.6$. $N=32$ terms. (Check: $\frac1{2\cdot31^2}\approx0.00052$ is too big; $\frac1{2\cdot32^2}\approx0.00049$ works.)
3 $\displaystyle\int_N^\infty x^{-4}dx=\frac{1}{3N^3}\le10^{-6}$ needs $N^3\ge333\,334$, $N\ge69.3$. $N=70$ terms ($69^3=328\,509$ is not enough; $70^3=343\,000$ is).
4 $f(x)=\frac1{x^2+1}$ and $\displaystyle\int_N^\infty\frac{dx}{x^2+1}=\frac\pi2-\tan^{-1}N$. So $S_{10}+\frac\pi2-\tan^{-1}11\le S\le S_{10}+\frac\pi2-\tan^{-1}10$: $$1.072453\le S\le1.081461.$$ (The exact sum is $\tfrac12(\pi\coth\pi-1)\approx1.076674$ — inside the bracket.)
5 $f(x)=xe^{-x^2}$ is decreasing on $[1,\infty)$ (Ex. II.3), so $R_3\le\displaystyle\int_3^\infty xe^{-x^2}dx=\tfrac12e^{-9}\approx6.2\times10^{-5}\lt 10^{-4}$. And $S_3=e^{-1}+2e^{-4}+3e^{-9}\approx0.36788+0.03663+0.00037=0.40488$.
6 Overhang $2$ means $H_n\ge4$. Enough: if $\ln(n+1)\ge4$, i.e. $n\ge e^4-1\approx53.6$, then certainly $H_n\ge4$ — so $54$ blocks are certainly enough. Needed: if $1+\ln n\lt 4$, i.e. $n\lt e^3\approx20.09$, then $H_n\lt 4$ — so at least $21$ blocks are needed. The bounds bracket the true answer $31$ ($H_{30}\approx3.995$, $H_{31}\approx4.027$). Because $H_n$ grows like $\ln n$, an overhang of $10$ block lengths would need about $e^{20}\approx5\times10^8$ blocks!
7 $\sum\frac{200}{n^{1.2}}=200\sum\frac1{n^{1.2}}$ is a constant times a $p$-series with $p=1.2\gt 1$, so the total is finite. With $\displaystyle\int_N^\infty 200x^{-1.2}dx=\frac{200N^{-0.2}}{0.2}=1000N^{-0.2}$, Theorem 8.14 with $N=30$ gives $$1000\cdot31^{-0.2}\le R_{30}\le1000\cdot30^{-0.2}\quad\Longrightarrow\quad 503.2\le R_{30}\le506.5,$$ so about $505$ aftershocks are predicted from day $31$ on (out of a total of about $1118$). With exponent $1$ the sum would be $200\sum\frac1n=\infty$: the model would predict infinitely many aftershocks, so it cannot be right for all time — real sequences are observed to have exponents slightly above $1$.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes
Research question. Word frequencies in a language roughly follow Zipf's law: the $r$-th most common word appears with frequency proportional to $1/r^s$. What does the integral test say about the exponent $s$, and how much of a text do the top $1000$ words cover?
Why it matters. Zipf-type laws appear in linguistics, city sizes, web traffic and earthquake magnitudes. Search engines, language-learning apps and text-compression software all rely on how quickly these sums settle.
What to do.
- If a language had infinitely many words with frequencies $C/r^s$, the frequencies must add up to $1$. Use Theorem 8.13 to explain why this forces $s\gt 1$, and why $s=1$ only works for a vocabulary of finite size $V$.
- For $s=1$ and $V=50\,000$ words, the share of the text covered by the top $1000$ words is $H_{1000}/H_{50000}$. Use the integral-test bounds $\ln(n+1)\le H_n\le1+\ln n$ to bracket this share, then compute it in a spreadsheet.
- Download a free word-frequency list (English, Arabic or another language you speak), plot $\log(\text{frequency})$ against $\log(\text{rank})$, and estimate $s$ from the slope.
- Repeat step 2 with your fitted $s$, using the remainder estimate to approximate the infinite tail.
Tools: Spreadsheet or Python; a public word-frequency list.
Hand in: A one-page note with your log–log plot, your estimate of $s$, the coverage of the top $1000$ words, and one reason real data departs from the model.
Going further. Compare two languages: does a language with richer word endings (such as Arabic or Finnish) have a flatter or steeper Zipf curve?
Instructor note: for $s=1$, $V=50\,000$: $H_{1000}\approx7.485$, $H_{50000}\approx11.397$, share $\approx65.7\%$; the crude bounds give roughly $58\%$ to $73\%$. Fitted exponents for large corpora are typically close to $1$.
🧪 Self-check quiz
0/5 correct
Q1. You know that $\displaystyle\lim_{n\to\infty}a_n=0$. What can you conclude about $\sum a_n$?
Q2. The series $\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}$
Q3. Which of these is not a hypothesis of the integral test for $\sum f(n)$?
Q4. For $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}$, the remainder estimate gives $R_N\le$
Q5. The series $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n(\ln n)^2}$