PG5.3 Absolute and conditional convergence

Simulate

Check the claim with numbers

The tests of lesson PG5.2 need terms that are never negative. Many series mix signs. The simplest is the alternating harmonic series, $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots$ , whose $k$ -th term is $\frac{(-1)^{k+1}}{k}$ .

Without the signs it is the harmonic series, which diverges (lesson PG5.1). With them, watch the partial sums $S_n$ .

import math import numpy as np k = np.arange(1, 10**6 + 1, dtype=float) signs = np.where(k % 2 == 1, 1.0, -1.0) S = np.cumsum(signs / k) print(" ".join(f"{s:.4f}" for s in S[:10])) for n in (10, 100, 1000, 10**6): print(n, round(float(S[n - 1]), 6)) print(round(math.log(2), 6)) even, odd = S[1::2], S[0::2] print(bool(np.all(np.diff(even) > 0)), bool(np.all(np.diff(odd) < 0)), bool(np.all(even < odd)))

⌘/Ctrl + Enter runs the cell

The first ten partial sums swing above and below a value near 0.69, ever closer. After a million terms the sum is 0.693147, which matches $\ln 2$ (lesson 2.7.2) to six decimals.

The last line checks three patterns for every $n$ up to a million. The even partial sums $S_2, S_4, \ldots$ go up, the odd ones $S_1, S_3, \ldots$ go down, and each $S_{2m}$ is below $S_{2m-1}$ .

Conjecture

\sum \frac{1}{k}

A million partial sums agree. The proof uses the three patterns the last line checked, for every $n$ .

Two kinds of convergence

The size of a term is $|a_k|$ . The series of sizes, $\sum |a_k|$ , has terms that are never negative, so the tests of lesson PG5.2 apply to it.

Rule: Absolute and conditional convergence

\begin{gathered} \textstyle\sum |a_k| \text{ converges:} \\ \text{absolutely convergent} \\[6pt] \textstyle\sum a_k \text{ converges and} \\ \textstyle\sum |a_k| \text{ does not:} \\ \text{conditionally convergent} \end{gathered}

A conditionally convergent series converges only because its signs cancel. The alternating harmonic series is the standard example, if the conjecture holds.

Format 2

Fill the gap

The proof of the conjecture, with its key steps left out. Type each missing piece as you would a quick answer, such as 1/(2m) or m+1, then check.

S_{2m}

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.

Simulate

The same terms in a new order

A finite sum gives the same total in any order. Try that on the alternating harmonic series: take each positive term, then the next two negative ones.

\begin{gathered} 1 - \tfrac{1}{2} - \tfrac{1}{4} \\ + \tfrac{1}{3} - \tfrac{1}{6} - \tfrac{1}{8} \\ + \tfrac{1}{5} - \tfrac{1}{10} - \tfrac{1}{12} + \cdots \end{gathered}

Every term of the old series appears exactly once. The cell adds the new series term by term, and checks the partial sums after $3m$ terms against $\frac{1}{2} S_{2m}$ exactly, with Fraction.

import math from fractions import Fraction def new_order(blocks): T, t = [], 0.0 for j in range(1, blocks + 1): for term in (1 / (2 * j - 1), -1 / (4 * j - 2), -1 / (4 * j)): t += term T.append(t) return T T = new_order(100_000) print(" ".join(f"{x:.4f}" for x in T[:9])) for n in (3, 30, 300, 3000, 300_000): print(n, round(T[n - 1], 6)) print(round(math.log(2), 6), round(math.log(2) / 2, 6)) def T_exact(m): return sum(Fraction(1, 2 * j - 1) - Fraction(1, 4 * j - 2) - Fraction(1, 4 * j) for j in range(1, m + 1)) def S_exact(n): return sum(Fraction((-1) ** (i + 1), i) for i in range(1, n + 1)) print(all(T_exact(m) == S_exact(2 * m) / 2 for m in range(1, 30)))

⌘/Ctrl + Enter runs the cell

The new partial sums settle near 0.3466, not 0.6931: after 300,000 terms they reach 0.346572, close to half of $\ln 2$ , 0.346574. The exact check holds for every $m$ up to 29.

Conjecture

\frac{L}{2}

The numbers say the order changed the sum. The next proof gets the reason wrong in one line; the other lines prove the new sum is $\frac{L}{2}$ .

Format 3

Find the error

A proof that $L = 0$ , which cannot be right: the proof above showed $L \ge \frac{1}{2}$ . One line is false. Select it.

Simulate

Signs that do not matter

When the sizes of the terms already add up, the signs cannot spoil convergence. Give each term of $\sum \frac{1}{k^2}$ a random sign, from a seeded coin, so the run repeats exactly.

import numpy as np rng = np.random.default_rng(5) k = np.arange(1, 10**6 + 1, dtype=float) a = rng.choice([-1, 1], size=k.size) / k**2 A = np.cumsum(a) for n in (10, 100, 1000, 10**6): print(n, round(float(A[n - 1]), 6)) print(round(float(np.abs(a).sum()), 6))

⌘/Ctrl + Enter runs the cell

The partial sums settle on 1.164916. The sizes add up to 1.644933 over the million terms, and the size of the sum, 1.164916, is below that.

Conjecture

\sum |a_k|

One seed, one sequence of signs. The proof must work for every choice of signs, and for any series whose sizes add up.

Format 4

Write it

Prove that if $\sum |a_k|$ converges, then $\sum a_k$ converges, with $\left|\sum a_k\right| \le \sum |a_k|$ . Hint: $b_k = a_k + |a_k|$ is never negative, and never above $2|a_k|$ .

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.

An absolutely convergent series can also be reordered freely: every order gives the same sum. This lesson states that without proof. Lebl's Basic Analysis, in the further reading below, proves it in chapter 2.

\sum a_k

Absolute and conditional convergence

Check the claim with numbers

Two kinds of convergence

Rule: Absolute and conditional convergence

Fill the gap

The same terms in a new order

Find the error

Signs that do not matter

Write it

Summary card

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