Absolute and Conditional Convergence; Rearrangement; Products

Let $(a_k)$ be a sequence of reals, with series $\sum a_k$ and partial sums $s_n = \sum_{k<n} a_k$ as in Series, partial sums, convergence and the sum, divergence, and the tail series, and let $|x|$ be the absolute value (Absolute value in an ordered field).

Absolute convergence. The series $\sum a_k$ converges absolutely when the series $\sum |a_k|$ converges (Series, partial sums, convergence and the sum, divergence, and the tail series). Since $|a_k| \ge 0$ for every $k$ (Basic properties of the absolute value), this is a statement about a series of nonnegative terms.

Conditional convergence. The series $\sum a_k$ converges conditionally when it converges (Series, partial sums, convergence and the sum, divergence, and the tail series, Limits and Cauchy sequences of reals) and does not converge absolutely.

So a convergent series is exactly one of the two: absolutely convergent or conditionally convergent, according as $\sum |a_k|$ converges or not.

One implication is already proved, and is not reproved anywhere on this page. If $\sum |a_k|$ converges then $\sum a_k$ converges states that if $\sum |a_k|$ converges then $\sum a_k$ converges. That lemma was coined and proved on the previous page of this track, where the root and ratio tests need it; this page names it and builds on it. In particular an absolutely convergent series is a convergent series, so the two words above really do partition the convergent series, and "conditionally convergent" is not vacuous by accident: the alternating harmonic series is a witness, and the witness is exhibited in FALSE: every convergent series converges absolutely.

General starting index. Let $m \in \mathbb{N}$ and let $(a_k)_{k \ge m}$ be a family from $m$ (Series, partial sums, convergence and the sum, divergence, and the tail series). The series $\sum_{k \ge m} a_k$ converges absolutely when $\sum_{k \ge m} |a_k|$ converges, and converges conditionally when it converges and does not converge absolutely. By Series, partial sums, convergence and the sum, divergence, and the tail series both statements are the corresponding statements for the shifted sequence $j \mapsto a_{j+m}$, so nothing new is being defined and every result below transfers to a general starting index in the same way, exactly as If $\sum |a_k|$ converges then $\sum a_k$ converges already records for the one implication it proves.

• Absolute convergence is a condition on the terms, not on the sum. It says the series of absolute values converges, and it says nothing about the value of $\sum_{k=0}^{\infty} a_k$. The two sums are in general different, and no statement here identifies them.

• Why the distinction earns a page. Every result on this page separates the two classes. An absolutely convergent series may be reordered at will (Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum) and multiplied by another (Mertens' theorem: if $\sum a_k$ converges absolutely to $A$ and $\sum b_k$ converges to $B$, their Cauchy product converges to $AB$); a conditionally convergent one may be reordered to any sum whatever (The Riemann series theorem: a conditionally convergent real series has, for every $c \in \mathbb{R}$, a rearrangement with sum $c$, and rearrangements diverging to $+\infty$, to $-\infty$, and oscillating with any prescribed $\liminf \le \limsup$ in $\overline{\mathbb{R}}$). The difference is not one of degree.

• A series of nonnegative terms converges absolutely if it converges at all, since then $|a_k| = a_k$. So the distinction is invisible for the comparison, condensation, Raabe, Gauss and Kummer tests of the previous page, all of which assume terms of one sign. It is not invisible on that page as a whole: the root and ratio tests are stated for terms of arbitrary sign and reach convergence of $\sum a_k$ precisely through If $\sum |a_k|$ converges then $\sum a_k$ converges, which is where the word absolutely convergent is first used. What that page does not develop, and this one does, is everything that separates the two classes rather than the one implication those two tests need.

Let $(a_k)$ be a sequence of reals (Series, partial sums, convergence and the sum, divergence, and the tail series) and define its positive part and negative part by

$a_k^{+} \;:=\; \frac{|a_k| + a_k}{2}, \qquad a_k^{-} \;:=\; \frac{|a_k| - a_k}{2} \qquad (k \in \mathbb{N}),$

with $|x|$ the absolute value (Absolute value in an ordered field). Then:

1. $a_k^{+} = \max\{a_k, 0\}$ and $a_k^{-} = \max\{-a_k, 0\}$ (Maximum and minimum of a set); in particular $a_k^{+} \ge 0$ and $a_k^{-} \ge 0$, and $a_k = a_k^{+} - a_k^{-}, \qquad |a_k| = a_k^{+} + a_k^{-} .$
2. $\sum a_k$ converges absolutely (Absolutely convergent and conditionally convergent series, and the general starting index) if and only if both $\sum a_k^{+}$ and $\sum a_k^{-}$ converge.
3. If $\sum a_k$ converges conditionally, then neither $\sum a_k^{+}$ nor $\sum a_k^{-}$ converges, and the partial sums of each diverge to $+\infty$ (Divergence to $+\infty$ and to $-\infty$).

Claim 3 is the engine of the rearrangement theory: a conditionally convergent series carries an unlimited supply of positive terms and an unlimited supply of negative ones, and its convergence is nothing but a cancellation between them.

• The two parts are determined by the terms, with no choice anywhere. The displayed formulas define $a^{+}$ and $a^{-}$ outright, and step 2.1 identifies them with the two maxima; nothing in the proof selects one of several candidates.

• Claim 3 is sharp in both directions. Absolute convergence makes both part series converge, and then $\sum_{k=0}^{\infty} a_k$ is the difference of their sums. Conditional convergence makes both part series diverge to $+\infty$, and the difference of their partial sums is what converges. There is no third possibility for a convergent series, because claim 2 covers the case where one of them converges: if exactly one converged, $\sum a_k = \sum(a_k^{+} - a_k^{-})$ could not converge, since the sum of a convergent and a divergent series diverges.

• Why $\max$ is mentioned at all. The formulas with $|a_k|$ are what the algebra uses, while $\max\{a_k, 0\}$ is what the name "positive part" means and what makes claims about signs immediate. Step 2.1 records that they agree, so either may be used later without further comment.

Let $(a_k)$ and $(b_k)$ be sequences of reals and let

$A_n \;:=\; \sum_{k<n} a_k \qquad (n \in \mathbb{N})$

be the partial sums of $\sum a_k$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion), so that $A_0 = 0$ and $a_k = A_{k+1} - A_k$ for every $k$. Then for every natural number $n \ge 1$

$\sum_{k<n} a_k b_k \;=\; A_n\, b_{n-1} \;-\; \sum_{k<n-1} A_{k+1}\,(b_{k+1} - b_k) .$

Both sides are finite sums in the sense of Finite sums and finite products, by recursion; at $n = 1$ the right-hand sum is empty and the identity reads $a_0 b_0 = A_1 b_0$.

The hypothesis $n \ge 1$ is what makes the statement legitimate, not merely convenient: the index $n-1$ occurs on the right, and $n-1$ is a natural number exactly when $n \ge 1$. At $n = 0$ there is nothing to state, both the left-hand side and $A_0$ being $0$.

• What the identity is for. It converts a series $\sum a_k b_k$, about which nothing is assumed, into a boundary term $A_n b_{n-1}$ and a series $\sum A_{k+1}(b_{k+1} - b_k)$ whose terms carry the differences of $(b_k)$. If $(A_n)$ is bounded and $(b_k)$ is monotone, those differences have one sign and telescope, which is exactly the situation of Dirichlet's test: if the partial sums of $\sum a_k$ are bounded and $(b_k)$ is nonincreasing with $b_k \to 0$, then $\sum a_k b_k$ converges. The transformation is the discrete analogue of integration by parts, and the boundary term is the analogue of the boundary term there.

• The block form needs no separate proof. For $1 \le M \le n$, subtracting the identity at $M$ from the identity at $n$ gives $\sum_{k=M}^{n-1} a_k b_k = A_n b_{n-1} - A_M b_{M-1} - \sum_{k=M-1}^{n-2} A_{k+1}(b_{k+1}-b_k)$, using only splitting of finite sums (Laws of finite sums and finite products). Nothing on this page needs that form, so it is recorded here rather than stated as a result.

• Two conventions are doing work. $A_n$ sums the $n$ terms $a_0, \dots, a_{n-1}$, so $A_0 = 0$ and $a_k = A_{k+1} - A_k$ with no shift (Series, partial sums, convergence and the sum, divergence, and the tail series); and the empty sum is $0$ (Finite sums and finite products, by recursion), which is what makes $n = 1$ a genuine instance of the identity rather than a case to be excluded.

Let $(a_k)$ and $(b_k)$ be sequences of reals, and let $A_n = \sum_{k<n} a_k$ be the partial sums of $\sum a_k$ (Series, partial sums, convergence and the sum, divergence, and the tail series). Suppose that

1. the range $\{\, A_n : n \in \mathbb{N} \,\}$ is bounded (Lower bound, bounded below, bounded set), that is there is a real $M \ge 0$ with $|A_n| \le M$ for every $n$; and
2. $(b_k)$ is nonincreasing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences) and converges to $0$ (Limits and Cauchy sequences of reals).

Then $\sum a_k b_k$ converges.

Under hypothesis 2 the terms $b_k$ are automatically nonnegative, and the proof says so before using it: a nonincreasing sequence is bounded below by each of its own later terms, and passing to the limit gives $b_k \ge 0$ (Limits preserve non-strict inequalities).

Nothing is assumed about $\sum a_k$ itself. Its partial sums need only stay bounded; they need not converge. That is what makes this test the source of the alternating series test (The alternating series test: if $(b_k)$ is nonincreasing with $b_k \to 0$ then $\sum_{k} (-1)^{k} b_k$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_n$) and of examples whose sign pattern is not alternating at all.

• Where each hypothesis is used, and none is decorative. Boundedness of $(A_n)$ is used twice: once to bound $|c_k|$ in step 2.1, and once to kill the boundary term in step 2.2. Monotonicity of $(b_k)$ is what makes $|b_{k+1} - b_k|$ equal to $b_k - b_{k+1}$, so that the bound in step 2.1 telescopes; without it the differences need not sum to anything. And $b_k \to 0$ is used both in the telescoping sum of step 1.4 and in the boundary term of step 2.2.

• Why nonincreasing and not monotone, although either would do. Hypothesis 2 could equally be stated with "monotone", and the theorem would still be true: a nondecreasing $(b_k)$ converging to $0$ is nonpositive, so $(-b_k)$ is nonincreasing and converges to $0$, and applying the theorem to it gives convergence of $\sum a_k(-b_k)$ and hence of $\sum a_k b_k$ (Convergent series add and scale termwise). What "monotone" may not be weakened to is "monotone and bounded": a monotone $(b_k)$ with a nonzero limit is not covered, and for such a factor the conclusion fails in general. The nonincreasing form is chosen here because it is the form the proof uses, and because it makes $b_k \ge 0$ immediate. Abel's test: if $\sum a_k$ converges and $(b_k)$ is monotone and bounded then $\sum a_k b_k$ converges is the result that handles monotone bounded factors, and it has a different hypothesis on $\sum a_k$.

• The sum is not computed. The proof produces the limit as $-T$, where $T$ is the sum of a series that the argument only proves convergent. This is a convergence test and nothing more.

Let $(\varepsilon_k)$ be the alternating sequence of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$, $s_{\sigma(k)} = -s_k$, which satisfies $|s_k| = 1$, $s \circ e \equiv 1$ and $s \circ o \equiv -1$, that is the unique sequence of reals with $\varepsilon_0 = 1$ and $\varepsilon_{k+1} = -\varepsilon_k$, which is what is usually written $\varepsilon_k = (-1)^k$; let $e$ and $o$ be its even and odd index maps, so that $\varepsilon_{e_j} = 1$, $\varepsilon_{o_j} = -1$, and every natural number is $e_j$ for exactly one $j$ or $o_j$ for exactly one $j$.

Let $(b_k)$ be a sequence of reals that is nonincreasing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences) and converges to $0$ (Limits and Cauchy sequences of reals); then $b_k \ge 0$ for every $k$. Write $t_n := \sum_{k<n} \varepsilon_k b_k$ for the partial sums (Series, partial sums, convergence and the sum, divergence, and the tail series). Then:

1. the series $\sum \varepsilon_k b_k$ converges; write $L$ for its sum;
2. $t_{e_j} \le L \le t_{o_j}$ for every $j \in \mathbb{N}$, and for every $n \in \mathbb{N}$ the sum $L$ lies between the two consecutive partial sums $t_n$ and $t_{n+1}$;
3. $|L - t_n| \le b_n$ for every $n \in \mathbb{N}$.

Claim 3 is the error bound: the partial sum $t_n$, which uses the $n$ terms $\varepsilon_0 b_0, \dots, \varepsilon_{n-1}b_{n-1}$, differs from the sum by at most the first term omitted.

Only claim 1 is a corollary of Dirichlet's test: if the partial sums of $\sum a_k$ are bounded and $(b_k)$ is nonincreasing with $b_k \to 0$, then $\sum a_k b_k$ converges. Claims 2 and 3 are not: they come from the interlacing of the even-index and odd-index partial sums, and that argument is carried out below rather than smuggled into the Dirichlet estimate, which produces no bracketing at all.