Absolute convergence

In mathematics, a series (or sometimes also an integral) of numbers is said to converge absolutely if the sum (or integral) of the absolute value of the summand or integrand is finite. More precisely, a real or complex series $\sum_{n=0}^\infty a_n$ is said to converge absolutely if $\sum_{n=0}^\infty \left|a_n\right| < \infty.$

Absolute convergence is important to the study of infinite series because its definition is strong enough to have properties of finite sums that not all convergent series possess, yet is broad enough to occur commonly.

Background

One may study the convergence of series $\sum_{n=0}^{\infty} a_n$ whose terms a_n are elements of an arbitrary abelian topological group. The notion of absolute convergence requires more structure, namely a norm:

A norm on an abelian group G (written additively, with identity element 0) is a real-valued function $x \mapsto \|x\|$ on G such that:

1. The norm of the identity element of G is zero: $\|0\| = 0.$
2. For every x in G, $\|x\| = 0 \implies x = 0.$
3. For every x in G, $\|-x\| = \|x\|.$
4. For every x, y in G, $\|x+y\| \leq \|x\| + \|y\|.$

Then the function $d(x,y) = \|x-y\|$ induces on G the structure of a metric space (in particular, a topology). We can therefore consider G-valued series and define such a series to be absolutely convergent if $\sum_{n=0}^{\infty} \|a_n\| < \infty.$

Relation to convergence

If the metric d on G is complete, then every absolutely convergent series is convergent. The proof is the same as for complex-valued series: use the completeness to derive the Cauchy criterion for convergence—a series is convergent if and only if its tails can be made arbitrarily small in norm—and apply the triangle inequality.

In particular, for series with values in any Banach space, absolute convergence implies convergence. The converse is also true: if absolute convergence implies convergence in a normed space, then the space is a Banach space.

If a series is convergent but not absolutely convergent, it is called conditionally convergent. An example of a conditionally convergent series is the alternating harmonic series. Many standard tests for divergence and convergence, most notably including the ratio test and the root test, demonstrate absolute convergence. This is because a power series is absolutely convergent on the interior of its disk of convergence.

Proof that any absolutely convergent series is convergent

Assume $\sum |a_n|$ is convergent. Since a series of complex numbers converges if and only if both its real and imaginary parts converge, we may assume with equal generality that $\forall n\ a_n \in \mathbb{R}$. Then, $2\sum |a_n|$ is convergent.

Since $0 \le a_n + |a_n| \le 2|a_n|$, we have:

$0 \le \sum_{n = 1}^m (a_n + |a_n|) \le \sum_{n = 1}^m 2|a_n|\le \sum_{n = 1}^\infty 2|a_n|$.

Thus, $\sum_{n = 1}^m (a_n + |a_n|)$ is a bounded monotonic sequence (in m), which must converge.

$\sum a_n = \sum(a_n+|a_n|) - \sum |a_n|$ is a difference of convergent series; therefore, it is also convergent. $\sum |a_n|$ is convergent $\to \sum a_n$ is convergent.

Rearrangements and unconditional convergence

In the general context of a G-valued series, a distinction is made between absolute and unconditional convergence, and the assertion that a real or complex series which is not absolutely convergent is necessarily conditionally convergent (meaning not unconditionally convergent) is then a theorem, not a definition. This is discussed in more detail below.

Given a series $\sum_{n=0}^{\infty} a_n$ with values in a normed abelian group G and a permutation σ of the natural numbers, one builds a new series $\sum_{n=0}^\infty a_{\sigma(n)}$, said to be a rearrangement of the original series. A series is said to be unconditionally convergent if all rearrangements of the series are convergent to the same value.

When G is complete, absolute convergence implies unconditional convergence.

Theorem

Let $\sum\limits_{i=0}^\infty a_i=A<\infty$, $\sum\limits_{i=0}^\infty |a_i|<\infty$ and let σ be a permutation of $\mathbb{N}$, then $\sum\limits_{i=0}^\infty a_{\sigma(i)}=A$

Proof

For anyε > 0, we can choose some $\kappa_\varepsilon,\lambda_\varepsilon \in \mathbb{N}$, such that

$\forall N>\kappa_\varepsilon\ ,\sum\limits_{n=N}^\infty ||a_n|| < \frac{\varepsilon}{2}$

and

$\forall N>\lambda_\varepsilon\ ,\left\|\sum\limits_{n=1}^N a_n-A\right\| < \frac{\varepsilon}{2}$.

let N_ε: = max(κ_ε,λ_ε), $M_{\sigma,\varepsilon}:= \max \left\{ \sigma^{-1}\left(\{ 1,\dots,N_\varepsilon \}\right) \right\}$.

For any $N > M_{\sigma,\varepsilon}\ \ (N \in \mathbb{N})$, let

$I_{\sigma,\varepsilon}:=\left\{ 1,\ldots,N \right\}\setminus \sigma^{-1}\left(\{ 1,\dots,N_\varepsilon \}\right)$,

$S_{\sigma,\varepsilon} := \min\{ \sigma(k) | k \in I_{\sigma,\varepsilon} \}$ (note.$\ S_{\sigma,\varepsilon} \geq N_{\varepsilon}+1$), and

$L_{\sigma,\varepsilon} := \max\{ \sigma(k) | k \in I_{\sigma,\varepsilon} \}$

then

$\left\|\sum\limits_{i=1}^N a_{\sigma(i)}-A \right\|= \left\| \sum_{i \in \sigma^{-1}\left(\{ 1,\dots,N_\varepsilon \}\right)} a_{\sigma(i)} - A + \sum_{i\in I_{\sigma,\varepsilon}} a_{\sigma(i)} \right\|$

$\leq \left\| \sum_{j=1}^{N_\varepsilon} a_j - A \right\| + \left\| \sum_{i\in I_{\sigma,\varepsilon}} a_{\sigma(i)} \right\| \leq \left\| \sum_{j=1}^{N_\varepsilon} a_j - A \right\| + \sum_{i\in I_{\sigma,\varepsilon}} \| a_{\sigma(i)} \|$

$\leq \left\| \sum_{j=1}^{N_\varepsilon} a_j - A \right\| + \sum_{j= S_{\sigma,\varepsilon} }^{ L_{\sigma,\varepsilon} } \| a_j \| \leq \left\| \sum_{j=1}^{N_\varepsilon} a_j - A \right\| + \sum_{j= N_\varepsilon + 1}^{\infty} \| a_j \| < \varepsilon$

therefore

$\forall\varepsilon > 0\ , \exist M_{\sigma,\varepsilon}\, \forall N > M_{\sigma,\varepsilon} \,\, , \left\|\sum\limits_{i=1}^N a_{\sigma(i)}-A \right\|< \varepsilon$

then $\sum\limits_{i=1}^\infty a_{\sigma(i)}=A$

Q.E.D.

The issue of the converse is much more interesting. For real series it follows from the Riemann rearrangement theorem that unconditional convergence implies absolute convergence. Since a series with values in a finite-dimensional normed space is absolutely convergent if each of its one-dimensional projections is absolutely convergent, it follows easily that absolute and unconditional convergence coincide for $\mathbb{R}^n$-valued series.

But there is an unconditionally and nonabsolutely convergent series with values in Hilbert space $\ell^2$: if $\{e_n\}_{n=1}^{\infty}$ is an orthonormal basis, take $a_n = \frac{1}{n} e_n$.

A theorem of Dvoretzky-Rogers asserts that every infinite-dimensional Banach space admits an unconditionally but non-absolutely convergent series.

Products of series

The Cauchy product of two series converges to the product of the sums if at least one of the series converges absolutely. That is, suppose:

$\sum_{n=0}^\infty a_n = A$

$\sum_{n=0}^\infty b_n = B.$

The Cauchy product is defined as the sum of terms c_n where:

$c_n = \sum_{k=0}^n a_k b_{n-k}.$

Then, if either the a_n or b_n sum converges absolutely, then

$\sum_{n=0}^\infty c_n = AB.$

Absolute convergence of integrals

The integral $\int_A f(x)\,dx$ of a real or complex-valued function is said to converge absolutely if $\int_A \left|f(x)\right|\,dx < \infty.$ One also says that f is absolutely integrable.

When A = [a,b] is a closed bounded interval, every continuous function is integrable, and since f continuous implies | f | continuous, similarly every continuous function is absolutely integrable. It is not generally true that absolutely integrable functions on [a,b] are integrable: let $S \subset [a,b]$ be a nonmeasurable subset and take $f = \chi_S - \frac{1}{2}$, where χ_S is the characteristic function of S. Then f is not Lebesgue measurable but | f | is constant. However, it is a standard result that if f is Riemann integrable, so is | f | . This holds also for the Lebesgue integral; see below. On the other hand a function f may be Kurzweil-Henstock integrable (or "gauge integrable") while | f | is not. This includes the case of improperly Riemann integrable functions.

Similarly, when A is an interval of infinite length it is well-known that there are improperly Riemann integrable functions f which are not absolutely integrable. Indeed, given any series $\sum_{n=0}^{\infty} a_n$ one can consider the associated step function $f_a: [0,\infty) \rightarrow \mathbb{R}$ defined by f_a([n,n + 1)) = a_n. Then $\int_0^{\infty} f_a dx$ converges absolutely, converges conditionally or diverges according to the corresponding behavior of $\sum_{n=0}^{\infty} a_n.$

Another example of a convergent but not absolutely convergent improper Riemann integral is $\int_{\mathbb{R}} \frac{\sin x}{x} dx$.

On any measure space A the Lebesgue integral of a real-valued function is defined in terms of its positive and negative parts, so the facts:

1. f integrable implies | f | integrable
2. f measurable, | f | integrable implies f integrable

are essentially built into the definition of the Lebesgue integral. In particular, applying the theory to the counting measure on a set S, one recovers the notion of unordered summation of series developed by Moore-Smith using (what are now called) nets. When $S = \mathbb{N}$ is the set of natural numbers, Lebesgue integrability, unordered summability and absolute convergence all coincide.

Finally, all of the above holds for integrals with values in a Banach space. The definition of a Banach-valued Riemann integral is an evident modification of the usual one. For the Lebesgue integral one needs to circumvent the decomposition into positive and negative parts with Daniell's more functional analytic approach, obtaining the Bochner integral.

See also

• Convergence of Fourier series
• Conditional convergence
• Modes of convergence (annotated index)
• Cauchy principal value
• A counterexample related to Fubini's theorem
• 1/2 − 1/4 + 1/8 − 1/16 + · · ·
• 1/2 + 1/4 + 1/8 + 1/16 + · · ·

References

• Walter Rudin, Principles of Mathematical Analysis (McGraw-Hill: New York, 1964).