A different definition of convergence of series

Question

I am reading the introduction to Hilbert space by P. R. Halmos. I found there the following definition of convergence of a series:

A family $\{x_j\}$ of vector will be called summable with sum $x$, in symbols $\sum\limits_{j} x_j=x$, if for every $\varepsilon > 0$ there exist a finite set $J_0$ of indices such that $\|x-\sum\limits_{j\in J} x_j \|< \varepsilon $, whenever $J$ is a finite set of indices containing $J_0.$

Now, my question is how this definition connects with the convergent of series in $\mathbb{R}$ or $\mathbb{C}$ (The definition involving convergence of the sequence of partial sum). If we assume this definition to be true, then we can always prove the definition of convergence of a series in $\mathbb{R}$ or $\mathbb{C}$. I can also prove the equivalence for convergent series of positive real numbers. However, I am finding it difficult to Equivalence in other cases.

Thanks in advance.

Answer 1

In $\mathbb{R}$ or $\mathbb{C}$, the statement is in fact equivalent to the absolute convergence of the series.

Let $(x_j)_{j\in\mathbb{N}}$ be a sequence in $\mathbb{C}$.

1. Suppose $\sum_{n=1}^{\infty} |x_n| < \infty$. For any $\varepsilon > 0$, choose $N$ such that $\sum_{n=N+1}^{\infty} |x_n| < \varepsilon$. Then with the choice $J_0 = \{1,2,\dots,N\}$, for any finite set $J$ with $J_0 \subseteq J \subseteq \mathbb{N}$,

$$ \left| \sum_{n=1}^{\infty} x_n - \sum_{j \in J} x_j \right| \leq \sum_{n=N+1}^{\infty} |x_n| < \varepsilon. $$

Therefore $\sum_j x_j$ is summable with the sum $\sum_{n=1}^{\infty} x_n$.

2. Suppose $\sum_j x_j$ is summable. Then for each $\varepsilon > 0$, we can pick $J_0$ such that

$$\left| \sum_{j \in J \setminus J_0} x_j \right| < \varepsilon/4$$

for any finite set $J$ with $J_0 \subseteq J \subseteq \mathbb{N}$. Also, by enlarging $J_0$ if necessary, assume that $J_0$ takes the form $J_0 = \{1, 2, \dots, N\}$ for some $N \in \mathbb{N}$. Now we define

\begin{align*} \mathsf{R}^+ &= \{ j \in \mathbb{N} : \operatorname{Re}(x_j) \geq 0 \}, & \mathsf{R}^- &= \{ j \in \mathbb{N} : \operatorname{Re}(x_j) < 0 \}, \\ \mathsf{I}^+ &= \{ j \in \mathbb{N} : \operatorname{Im}(x_j) \geq 0 \}, & \mathsf{I}^- &= \{ j \in \mathbb{N} : \operatorname{Im}(x_j) < 0 \}. \end{align*}

Then for any $M > N$,

$$ \sum_{j \in (N, M] \cap \mathsf{R}^+} \left| \operatorname{Re}(x_j) \right| = \operatorname{Re} \Biggl[ \sum_{j \in (N, M] \cap \mathsf{R}^+} x_j \Biggr] \leq \left| \sum_{j \in (N, M] \cap \mathsf{R}^+} x_j \right| < \varepsilon/4, $$

and similar arguments show that

$$ \sum_{j \in (N, M] \cap \mathsf{R}^-} \left| \operatorname{Re}(x_j) \right| < \varepsilon/4, \qquad \sum_{j \in (N, M] \cap \mathsf{I}^+} \left| \operatorname{Im}(x_j) \right| < \varepsilon/4, \qquad \sum_{j \in (N, M] \cap \mathsf{I}^-} \left| \operatorname{Im}(x_j) \right| < \varepsilon/4. $$

So it follows that

$$ \sum_{j=N+1}^{M} |x_j| \leq \sum_{j=N+1}^{M} (\left| \operatorname{Re}(x_j) \right| + \left| \operatorname{Im}(x_j) \right|) < \varepsilon, $$

and this proves that $\sum_{j=1}^{\infty} |x_j| < \infty$. Then the previous step shows that $\sum_{j=1}^{\infty} x_j = \sum_j x_j$.

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Remarks.

• By extending the above proof, we can also prove that the followings are equivalent for any families $\{x_j\}_{j\in\mathcal{J}}$ in $\mathbb{C}^n$:

  1. $\{x_j\}_{j\in\mathcal{J}}$ is summable.

  2. $\{\|x_j\|\}_{j\in\mathcal{J}}$ is summable.

  3. $\sup_{J} \sum_{j \in J} \|x_j\| < \infty$, where the supremum is taken over all finite subsets $J$ of $\mathcal{J}$.

• On the other hand, the equivalence fails in infinite-dimensional vector spaces. For example, consider the family $\{x_j\}_{j\in\mathbb{N}}$ given by $x_j = (j^{-1} \delta_{jn})_{n=1}^{\infty}$ in the space $\ell^2(\mathbb{N})$ of the square-summable sequences. This family is summable but not "absolutely summable".

Answer 2

If $\sum a_n$ is convergent in $\mathbb R$ or $\mathbb C$ it does not folow that the series is summable in above sense. Consider $\sum \frac {(-1)^{n}} n$. Let $s$ be its sum. Whatever finite set $I$ of integers you choose you can find a finite set $J$ containing $I$ such that $|s-\sum_{j \in J} a_j|$ is as large as you want : Just take all even intgers from the maximum of $I$ up to some large integer.

However if $\sum |a_n| <\infty$ the $\sum a_n$ is summable in above sense.

Answer 3

Here is a another solution that shows the following:

Proposition 1: Suppose $I$ an non empty set, and denote by $\mathcal{F}(I)$ the family of finite subsets of $I$. Let $x:I\rightarrow\mathbb{R}$ be a function.

1. If $x(j)\geq 0$ for all $j$, then the net $S(J)=\sum_{j\in J}x(j)$ converges iff $\{S(J):J\in\mathcal{F}(I)\}$ is bounded. In such case, $$s=\sup\{S(J): J\in\mathcal{F}(I)\}$$
2. If $x$ is not necessarily nonnegative, then the net $S(D)$ converges iff the net $S^a(J)=\{\sum_{J\in \mathcal{F}}|x(j)|: J\in\mathcal{F}(I)\}$ converges. In this case $s:=\lim_JS(J)\leq \lim_jS^a(J)=:s^a$.

Two direct consequences of this results are

Corollary A. Under the assumptions of Proposition 1, for any function $x:I\rightarrow\mathbb{C}$, the complex net $\{S(J)=\sum_{j\in J}x(j):J\in\mathcal{F}(I)\}$ converges iff the nonnegative net $\{S^a(J)=\sum_{j\in J}|x(j)|:J\in\mathcal{F}(I)\}$ converges.

Proof: Suppose $a+ib=s=\lim_{J\in\mathcal{F}(I)}S(J)$. Let $S^r(J)=\operatorname{Re}(S(J))$ and $S^i(J)=\operatorname{Im}(S(J))$. The conclusion follows from Proposition 1 and the well known relation between a complex number and its real and imaginary parts $$\max(|a-S^r(J)|,|b-S^I(J)|)\leq |S(J)-s|\leq |a-S^r(J)|+|b-S^I(J)|\leq \sqrt{2}|s-S(J)|$$

Corollary B. If $I=\mathbb{N}$, then the net (real of complex) $S(J)$, $J\in\mathcal{F}(\mathbb{N})$ converges iff the series $\sum^\infty_{n=1}x(n)$ converges absolutely

Proof: By Proposition 1 and Corollary A it suffices to assume that $x$ is nonnegative. In this case, Proposition 1 states that the net $S(J)$, $J\in\mathcal{G}(\mathbb{N})$ converges iff $\{S(J):J\in\mathcal{F}(\mathbb{N})\}$ is bounded. This happens iff $\{s_n=\sum^n_{j=1}x(j):n\in\mathbb{N}\}$ is bounded, and this iff $\sum^\infty_{n=1}x(j)$ converges.

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The rest of this posting concerns the proof of Proposition 1.

(1) Suppose $x\geq0$. If $S(J)$ converges, then for any $\varepsilon>0$ there is $J_\varepsilon\in\mathcal{F}$ such that $|S(J)-s|<\varepsilon$ whenever $J\in\mathcal{F}$ and $J_\varepsilon\subset J$; hence $0\leq S(J)\leq s+\varepsilon$ for all $J\in\mathcal{F}$ with $J_\varepsilon\subset J$. This means that $\alpha:=\sup\{S(J):J\in\mathcal{F}\}\in[0,\infty)$ which in turn implies that $s\leq \alpha$. For any $J\in\mathcal{F}$ we have $0\leq S(J)\leq S(J\cup J_\varepsilon)<s+\varepsilon$. Hence $\alpha\leq s+\varepsilon$, and since $\varepsilon>0$ is arbitrary, $\alpha\leq s$. Therefore $s=\alpha$.
Conversely, suppose $\alpha=\sup\{S(J):J\in\mathcal{F}\}<\infty$. Given $\varepsilon>0$, choose $J_\varepsilon$ such that $\alpha-\varepsilon<S(J_\varepsilon)$. Since $x\geq0$, for any $J_\varepsilon\subset J$ $$\alpha-\varepsilon<S(J_\varepsilon)\leq S(J)\leq\varepsilon<\alpha+\varepsilon$$ Hence $S(J)$ converges to $\alpha$.

(2) (sufficiency) Suppose $S^a(J)$ converges to some $s^a\geq0$. Then, there exists a sequence of sets $J_n\in\mathcal{F}$ such that $J_n\subset J_{n+1}$ and $|s-S^a(J)|<2^{-n}$ for all $J_n\subset J$. By the first part of the present Lemma, For any $n>m$ $$|S(J_n)-S(J_m)|\leq S^a(J_n)-S^a(J_m)\leq s^a-S^a(J_m)<2^{-m}$$ which means that $\{S(J_n):n\in\mathbb{N}\}$ is a Cauchy sequence; hence $s=\lim_nS(J_n)$ exists and is finite. Moreover, $$|s-S(J_m)|=\lim_n|S(J_n)-S(J_m)|\leq\lim_n|S^a(J_n)-S^a(J_m)|\leq 2^{-m}$$ Then, if $J\in\mathcal{F}$ and $J_n\subset J$, $$|s-S(J)|\leq |s-S(J_n)|+|S(J_n)-S(J)|\leq 2^{-n}+S^a(J)-S^a(J_n)\leq 2^{-n}+2^{-n}<2^{-n+1}$$ We conclude that $S(J)$ converges to $s$.

(necessity) Suppose that $S(J)$ converges to some $s$. Let $I_+=\{j\in I: x(j)\geq0\}$ and $I_-=I\setminus I_+$. We claim that $S^+:\mathcal{F}(I_+)\rightarrow[0,\infty)$ and $S^-:\mathcal{F}(I_-)\rightarrow[0,\infty)$ given by $S^+(J')=\sum_{j\in J'}x(j)$ and $S^-(J'')=-\sum_{j\in J''}x(j)$ converge. Fix $J_0\in\mathcal{F}$ such that $|s-S(J)|<1$ whenever $J^0\subset J$.
Let $J^0_+=J^0\cap I_+$ and $J^0_-=J^0\setminus J^0_+$. Then, for any $J'\in\mathcal{F}(I_+)$ and $J''\in\mathcal{F}(I_-)$ \begin{align*} S^+(J^0_+\cup J')&=|S(J^0\cup J')+S(J^0_-)-s+s|\leq |s|+\sum_{j\in J^0_-}|x(j)|+1\\ S^-(J^0_-\cup J'')&=|S(J^0\cup J'')-S(J^0_+)-s+s|\leq |s|+\sum_{j\in J^0_+}|x(j)|+1 \end{align*} As both $J^0_+$ and $J^0_-$ are finite, we conclude that $\{S^+(J'): J'\in\mathcal{F}(I_+)\}$ and $\{S(J''):J''\in\mathcal{F}(I_-)\}$ are bounded; hence, by part (1) both $S^+(J')$ and $S^-(J'')$ converge, say to $s^+$ and $s^-$ respectively. Consequently, for $\varepsilon>0$, there are $J^\pm_\varepsilon\in\mathcal{F}(I_\pm)$ such that whenever $J^+_\varepsilon\subset J'$ and $J^-_\varepsilon\subset J''$, $$|s^+-S^+(J')|<\frac{\varepsilon}{2},\qquad|s^--S^-(J'')|<\frac{\varepsilon}{2}$$ Let $J_\varepsilon=J^+_\varepsilon\cup J^-_\varepsilon$. It is clear that for any $J$ containing $J_\varepsilon$, $J'=J\cap I_+\supset J^+_\varepsilon$ and $J''=J\cap I_-\supset J^-_\varepsilon$. Hence \begin{align*} |s^+-s^--S(J)|=|s^+-S^+(J')+S^-(J'')|\leq |s^+-S^+(J')|+|s^--S^-(J'')|<\frac{\varepsilon}{2} \end{align*} whenever $J_\varepsilon\subset J$.