PG5.2 The comparison and ratio tests, proved

Simulate

Check the claim with numbers

Lesson PG5.1 showed that a series needs more than shrinking terms. This lesson proves two tests that can decide. The first compares a series with one whose sum is known.

Take $a_k = \frac{|\sin k|}{2^k}$ , with $k$ in radians. Its terms are irregular, but never negative and never above $b_k = \frac{1}{2^k}$ , whose series adds up to 1 (lesson 1.5.3). The cell prints both partial sums.

import math A = B = 0.0 for k in range(1, 61): A += abs(math.sin(k)) / 2**k B += 1 / 2**k if k in (1, 2, 5, 10, 20, 40, 60): print(k, round(A, 10), round(B, 10))

⌘/Ctrl + Enter runs the cell

The partial sums of $\sum a_k$ only go up, and they stay below those of $\sum b_k$ . They settle: after 40 and after 60 terms the sum is 0.7584330388, the same to ten decimal places.

Conjecture

0 \le a_k \le b_k

One pair of series, sixty terms. The proof must work for every pair, with no formula for the sum of $\sum a_k$ .

Series with terms that are never negative

When every $a_k \ge 0$ , each new term can only raise the partial sum: $S_n = S_{n-1} + a_n \ge S_{n-1}$ . So the partial sums form an increasing sequence, and two facts about limits apply.

Increasing and bounded. An increasing sequence that is bounded above converges (lesson PG4.5).
Limits keep inequalities. If $x_n \to L$ and $x_n \le c$ from some point on, then $L \le c$ . Otherwise $\varepsilon = L - c$ would put every $x_n$ from some later point on within $L - c$ of $L$ , so above $c$ . The same holds with $\ge$ in place of $\le$ .

Rule: Nonnegative terms

\begin{gathered} a_k \ge 0 \text{ for every } k, \\ S_n \le B \text{ for every } n \\ \Longrightarrow \sum_{k=1}^{\infty} a_k \text{ converges} \\ \text{and its sum is at most } B \end{gathered}

The partial sums go up and never pass $B$ , so they converge, and their limit is at most $B$ . Here $B$ is any number that no partial sum exceeds.

Format 1

Reorder the steps

Here is the proof of the comparison test, with its lines shuffled. Put them in order with the arrows, or drag a line by its handle.

Every $b_k \ge 0$ , so $B_n \le B_m$ whenever $n \le m$ . Letting $m \to \infty$ , limits keep the inequality: $B_n \le B$ for every $n$ .
Since $a_k \le b_k$ for every $k$ , adding the first $n$ of each gives $A_n \le B_n \le B$ .
Since $A_n \le B$ for every $n$ , the limit keeps the inequality: $\sum a_k = A \le B = \sum b_k$ .
Write $A_n$ and $B_n$ for the partial sums of $\sum a_k$ and $\sum b_k$ , and $B$ for the sum of $\sum b_k$ .
Every $a_k \ge 0$ as well, so the $A_n$ only go up, and the line before says they never pass $B$ .
An increasing sequence that is bounded above converges (lesson PG4.5), so $A_n$ tends to a limit $A$ .

Read backwards, the test also proves divergence. If $0 \le a_k \le b_k$ and $\sum a_k$ diverges, then $\sum b_k$ diverges too, since otherwise the test would make $\sum a_k$ converge.

Simulate

A series to compare

Lesson 1.5.3 found the partial sums of $1 + \frac{1}{4} + \frac{1}{9} + \cdots$ levelling off near 1.644934. To prove that they converge, compare each term from $k = 2$ on with $\frac{1}{k(k-1)}$ , which is a little larger.

from fractions import Fraction import numpy as np k = np.arange(1, 10**6 + 1, dtype=float) S = np.cumsum(1 / k**2) for n in (10, 100, 1000, 10**6): print(n, round(S[n - 1], 6), round(2 - 1 / n, 6)) print(all(Fraction(1, k**2) <= Fraction(1, k * (k - 1)) for k in range(2, 1000))) print(sum(Fraction(1, k - 1) - Fraction(1, k) for k in range(2, 11)))

⌘/Ctrl + Enter runs the cell

Each partial sum $S_n$ stays below $2 - \frac{1}{n}$ : 1.549768 against 1.9 for $n = 10$ , and 1.644933 against 1.999999 for a million terms. The comparison holds exactly from $k = 2$ to 999. The last line adds $\frac{1}{k-1} - \frac{1}{k}$ from $k = 2$ to 10: almost everything cancels, leaving $\frac{9}{10}$ .

Conjecture

\sum_{k=1}^{\infty} \frac{1}{k^2}

Checked up to a million terms. The proof gives the bound $2 - \frac{1}{n}$ for every $n$ .

Format 2

Fill the gap

The proof, with its key steps left out. Type each missing piece as you would a quick answer, such as 1/(k+1) or 3-2/n, then check.

k \ge 2

How to type answers

Powers with ^, brackets round a power that is a sum: r^k, r^(k+1).
Fractions with / and brackets: (1-r)/(1+r). Multiply with * or side by side: 2k+1.
Press Enter in any box to check them all.

Lesson 1.5.3 gave the sum: $\frac{\pi^2}{6} \approx 1.644934$ . The proof shows only that a sum exists, below 2. That is often all a proof of convergence gives.

Simulate

The ratio of one term to the next

The second test needs no partner series. For positive terms, it looks at the ratio $\frac{a_{k+1}}{a_k}$ of each term to the one before. The cell prints the ratio at $k = 1$ , 10 and 100, then the partial sums after 100 and 1,000 terms.

from math import factorial rows = { "k/2^k": lambda k: k / 2**k, "3^k/k!": lambda k: 3**k / factorial(k), "1/k": lambda k: 1 / k, "1/k^2": lambda k: 1 / k**2, } for name, a in rows.items(): ratios = [round(a(k + 1) / a(k), 4) for k in (1, 10, 100)] S = [round(sum(a(k) for k in range(1, n + 1)), 4) for n in (100, 1000)] print(f"{name:>7}", ratios, S) print(all((k + 1) / (2 * k) > 0.5 for k in range(1, 10**5)))

⌘/Ctrl + Enter runs the cell

For $\frac{k}{2^k}$ the ratios close in on 0.5, and the partial sums settle on 2.0. For $\frac{3^k}{k!}$ they close in on 0, and the sums settle on 19.0855. Ratios that end up below 1 behave like a geometric series.

The last line shows a detail that matters below. The ratios of $\frac{k}{2^k}$ are $\frac{k+1}{2k}$ , and all of them are above 0.5, up to $k = 10^5$ .

Both $\frac{1}{k}$ and $\frac{1}{k^2}$ have ratios that close in on 1. The first series diverges (lesson PG5.1) and the second converges, so a limit of 1 decides nothing.

Conjecture

a_k > 0

Four series, three ratios each, and the ratios of $\frac{k}{2^k}$ up to $k = 10^5$ . The proof compares the series with a geometric series, which lesson 1.5.3 summed.

The proof uses one more fact. Leaving out the first $N$ terms of a series moves every partial sum by the same amount, $a_1 + \cdots + a_N$ . So it does not change whether the series converges.

Format 3

Find the error

A proof of the ratio test, with one line that does not follow from what comes before it. Select it.

Simulate

Ratios above 1

Take $a_k = \frac{2^k}{k^3}$ . Its ratios tend to 2, above 1. The cell prints $k$ , the term $a_k$ and the ratio $\frac{a_{k+1}}{a_k}$ .

⌘/Ctrl + Enter runs the cell

The terms fall at first, to 0.25 at $k = 4$ , while the ratios are below 1. In the rows from $k = 4$ on, the ratios are above 1 and the terms grow: 1.024 at $k = 10$ , 131.072 at $k = 20$ .

Conjecture

a_k > 0

Nine terms. The proof turns a ratio above 1 into terms that cannot shrink to 0.

Format 4

Write it

Prove that if every $a_k > 0$ and $\frac{a_{k+1}}{a_k} \to L$ with $L > 1$ , then $\sum a_k$ diverges. Choose an $\varepsilon$ , then use the term test of lesson PG5.1.

Your proof

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Write your own first: a proof you have tried to write teaches far more than one you only read.

Lesson PG5.4 uses the ratio test on the series for $e^x$ . For every $x \ne 0$ , the ratios of the sizes of its terms tend to 0.

B

The comparison and ratio tests, proved

Check the claim with numbers

Series with terms that are never negative

Rule: Nonnegative terms

Reorder the steps

A series to compare

Fill the gap

The ratio of one term to the next

Find the error

Ratios above 1

Write it

Summary card

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