# What can be computed by axiomatic summation?

Here are three simple properties one might require of a summation method for divergent series:

• A stable summation scheme is one in which (assuming also each sums are defined iff the other is) $$\sum_{n=1}^\infty a_n = a_1 + \sum_{n=1}^\infty a_{n+1}$$
• A linear one satisfies (whenever the RHS is defined) $$\sum_{n=1}^\infty (\kappa a_n+b_n) = \kappa\sum_{n=1}^\infty a_n + \sum_{n=1}^\infty b_n$$(Edit: Added $\kappa$, that was forgetfulness.)
• A regular one agrees with normal summation when both are defined. (That is, if one has a convergent series, it is given its normal value by the summation method.)

If one assumes these rules, what class of series may be computed? It sounds like it should be the set of series which differ by a convergent series from another which is periodic up to rescaling; i.e. $a_n = c_n + k^n p_n$ where $\sum c_n$ is convergent and $p_{n+m} = p_n$ for some $m$. Is that correct? (Edit: Clearly this guess should have been $c_n + \sum_{i=1}^M k_{(i)}^n p_{(i) n}$, thanks Karene. Further, I was already aware of the issue whereby stability gives contradictions to e.g. $1+1+\cdots$ so my conjecture should also have reflected that.)

Given a specific, arbitrary series not of this form, can one actually/necessarily either (a) find two stable, linear, regular summation schemes for which the sum is defined, but for which we get distinct results; or (b) prove that no such scheme can define a value for the sum? (Motivation: pure curiosity.)

• Could you make your question more clear? What do you exactly mean by regular summation? – dani_s Jan 11 '14 at 7:33
• According to Wikipedia, yes, there exist different stable, linear and regular summation methods that can give different results for the same series. Unfortunately I don't have a specific example to give, although apparently Abelian means with different $\lambda$ can provide one. – Ilmari Karonen Jan 11 '14 at 10:02
• You want your conjecture on what series could be summed to be closed by the properties of your summation method. If you assign a sum to two sequences $c_n+k^np_n$, $c_n+r^np_n$, with $p_n$ periodic, you also assign a sum to $2c_n+k^np_n+r^np_n$, which is not of the form you propose. – user119256 Jan 11 '14 at 11:09
• To get a form of series with values determined by the axioms I would look for solutions to linear recurrence relations, with constant coefficients, $C_ka_{n+k}+C_{k-1}a_{n+k-1}+...+C_0a_n=f_n$, with independent term $f_n$ such that $\sum f_n$ is convergent (or summable), and with $C_0+C_1+...+C_{k}\neq0$. If your linearity (additivity rather) is not also full linearity, then maybe consider only $C_i$ rational numbers. – user119256 Jan 11 '14 at 11:33
• @dani_s & Karene, thanks, fixed. Also restored linearity to its intended definition. – Sharkos Jan 11 '14 at 17:13

I will assume that linearity is $$\sum(\alpha a_n+\beta b_n)=a\sum a_n+\beta\sum b_n.$$ If not, the constant coefficients below take them to be rational.

Denote by $S(a_1,a_2,...):=(0,a_1,a_2,...)$ the unilateral shift operator. Denote by $A$ the summation functional, that takes a series and returns its sum. We assume that $A$ is stable, linear, and regular.

We have that $A\circ S=A$, from stability. Moreover, stability is equivalent (assuming also linearity) to

$$A\circ S=A.$$

In fact, if $A\circ S=A$, and since $S(a_1,0,0,...)=a_1$,

\begin{align}A(a_1,a_2,...)&=A(a_1,0,0,...)+A(0,a_2,a_3,...)\\&=a_1+A(0,a_2,...)\\&=a_1+A(a_2,a_3,...)\end{align},

which is the axiom of stability.

If $A(\vec{a})$, with $\vec{a}:=(a^{(1)},a^{(2)},...)$ a vector of sequences, and $a^{(i)}:=(a_{i1},a_{i2},...)$, can be determined by the axioms it means that there is a finite number of applications of the axioms to the sequence and to some convergent series that gives you an equation from which the sum id determined.

Denote by $L_C:=C_kS^k+C_{k-1}S^{k-1}+...+C_1S+C_0I$ to be a linear finite-difference operator with constant coefficients, where the $C_i$ are vectors and the $S$ is applied to vectors of sequences componentwise. Then we can translate the above statement on being determined by the axioms into:

There exist finitely many vectors of convergent series $\sum f_n^{(1)},\sum f_n^{(2)},...,\sum f_n^{(r)}$, and linear operators $L_{C^{(1)}}, L_{C^{(3)}},...,L_{C^{(r)}}$, and $L_C$, such that

$$L_C(a)=L_{C^{(1)}}(f^{(1)})+L_{C^{(3)}}(f^{(2)})+...+L_{C^{(r)}}(f^{(r)}),$$

and $d:=\det[C_k|C_{k-1}|...|C_0]\neq0$.

This is what allows to apply $A$ in both sides and obtain $$dA(\vec{a})=A(L_{C^{(1)}}(f^{(1)})+L_{C^{(3)}}(f^{(2)})+...+L_{C^{(r)}}(f^{(r)})),$$ from where the sum is computed.

Clearly the sequence in the right-hand side is convergent. Therefore we get that $$L_C(\vec{a})=f$$ for some $f:=(f_1,f_2,...)$, with $\sum f_n$ convergent (usual sense).

This answers your first question (as I understand it) about what series' values are determined by the axioms:

Answer: These solutions of linear recurrence equations with constant coefficients adding up to a non-zero number, and with independent term being the term of a convergent series (the usual convergence).

You second question about different summations satisfying the axioms and defining sums in incompatible ways seems to be answered in the positive in Wikipedia (as dani_s pointed out). Maybe later I (or someone can give an explicit example of that).

It remains though the following

Question: Consider the series that have sums determined by the axioms (those solutions of recurrences). Are the values determined by the axioms consistent? In other words, is it possible to have a series satisfying two of these recurrence equations that then determine for the series two different values for its generalized sum?

• It seems the following. The series which we can use in the derivations of the axioms should belong to $dom A$. Your reasoning seems to be suitable for the case $dom A=\langle \mathcal C\cup \{S^k(a):k$ is a non-negative integer $\} \rangle$. But if we generate $dom A$ as a span using several sequences, we may obtain a system of recurrence equations $L_{C_1}(a,b,c,\dots z)=f_1,$ $L_{C_2}(a,b,c,\dots z)=f_2,\dots$ instead of one equation $L_C(a)=f$. – Alex Ravsky Jan 11 '14 at 14:52
• I expect a positive answer to Question, that is the values are consistent. – Alex Ravsky Jan 11 '14 at 14:54
• @AlexRavsky You are right. From linearity we get not just linear equations but linear systems. – user119256 Jan 11 '14 at 15:13
• Thanks; as to the second question, notice I asked about being given an arbitrary (specific) series - so can one choose two $\lambda_n,\lambda'_n$ in the Abelian mean to give different results? Probably this is easy, but my brain is slightly melted at the moment. – Sharkos Jan 11 '14 at 17:33
• @Sharkos It depends on the series. If it is a solution of these recurrences then you cannot give an arbitrary value. For the rest Ravky's answer tells you you can. – user119256 Jan 11 '14 at 23:20

It seems the following.

A linear summation scheme is a linear operator on a subspace of the linear space $\mathcal S$ of all series. The continuity of a summation scheme is not necessary, so this is mainly a question from linear algebra, not from real analysis.

Everywhere below $\Sigma$ shall be a stable, linear, and regular summation scheme defined on a subspace $\mathcal S_\Sigma$ of the linear space $\mathcal S$. Clearly, $\mathcal S_\Sigma$ is a proper subspace of the space $\mathcal S$, because in the opposite case we would have $\Sigma 1=1+\Sigma 1$, where $1=\{1\}_n$ is a constant sequence. Similarly we can show that if $a$ is an eventually periodic series and $\Sigma a$ is defined then the members of a (finite) periodic part of $a$ should sum up to the zero.

Moreover, $\Sigma$ may have counter intuitive properties. For instance, if $\{q^{n-1}\}\in \mathcal S_\Sigma$ is a geometric series then $\Sigma \{q^{n-1}\}=1+q\Sigma \{q^{n-1}\}$ so $\Sigma \{q^{n-1}\}=\frac 1{1-q}$. In particular, $\Sigma \{2^{n-1}\}=-1$.

Let $\mathcal C$ be a linear space of all convergent series and $\Pi:\mathcal S\to \mathcal S/\mathcal C$ be the quotient homomorphism. Following Karene, I denote by $S(a_1,a_2,\dots):=(0,a_1,a_2, \dots)$ a unilateral shift operator on the space $\mathcal S$. Since $\mathcal C$ is an invariant subspace of the operator $S$, there exists a linear operator $S’$ on the quotient space $\mathcal S/\mathcal C$ such that $\Pi S’=S\Pi$. For an arbitrary series $a\in\mathcal S$ put $\mathcal C(a)=\langle \mathcal C\cup \{S^k(a):k$ is a non-negative integer $\} \rangle$ and $\mathcal C’(a)=\Pi\mathcal C(a)$. The space $\mathcal C’(a)$ is invariant for the operator $S’$. If the space $\mathcal C’(a)$ is finitely-dimensional (that is, when $\mathcal C$ is a linear subspace of finite codimension in $\mathcal C(a)$) then the operator $S’|\mathcal C’(a)$ has a minimal polynomial which we denote as $p_a$.

Conjecture 1. Let $a\in\mathcal S$ be a series. The following conditions are equivalent:

1. A sum $\Sigma a$ is determined by the axioms.

2. A sum $\Sigma a$ is uniquely determined by the axioms.

3. The series $a$ is a solution of a linear recurrence $L_C(a)=f$ for some $f\in\mathcal C$ ($L_C$ is defined in Karene’s answer. In fact, $L_C$ is $l(S)$, where $l$ is a polynomial).

4. The space $\mathcal C’(a)$ is finitely dimensional and $p_a(1)\ne 0$.

As a partial result I prove that 2 implies the space $\mathcal C’(a)$ is finitely dimensional. Indeed, assume the converse. Then the family of series $\{\Pi S^k(a):k$ is a non-negative integer $\}$ is linearly independent. We extend a linear operator of the normal summation from the linear space $\mathcal C$ to a linear operator $\Sigma$ defined on the space $\mathcal C(a)$ as follows. We can define $\Sigma$ by putting $\Sigma a=\alpha$, where $\alpha$ is an arbitrary number and then continuing recurrently on the sequence $S^k(a)$ as $\Sigma S^{k+1}a=\Sigma S^{k}a-a_k$, where $a=\{a_k\}$. Then we extend $\Sigma$ linearly to the whole linear space $\mathcal C(a)$.

Conjecture 2. Let $a\in\mathcal S$ be a series. The following conditions are equivalent:

1. A sum $\Sigma a$ cannot be defined.

2. The space $\mathcal C’(a)$ is finitely dimensional, $p_a(1)=0$ and the sequence $p_a(S)a$ does not converge to zero.

Remark. I think that if the series $a$ grows faster than a geometric series then the space $\mathcal C’(a)$ is infinitely dimensional. I think that the proof of this is easy and leave it to an interested reader.

• It seems to be there are examples (some Abelian means) of non-consistent, stable, linear, and regular, but it is not as simple as choosing a divergent series and giving it a sum. There are some hidden conditions that should also be satisfied. For example, the summation method cannot assign a sum to divergent series such that for some $k$ the sequence of sums of $k$ consecutive of its terms has a limit. You also showed that some values of sums are forced. If a geometric series is summed it must have the sum you computed. – user119256 Jan 11 '14 at 11:03
• Above I meant to say "... of its terms has a limit different from zero." – user119256 Jan 11 '14 at 11:10
• Solutions to linear recurrence relations, with constant coefficients, $C_ka_{n+k}+C_{k-1}a_{n+k-1}+...+C_0a_n=f_n$, with independent term $f_n$ such that $\sum f_n$ is convergent (or summable) will have forced sums if a sum is to be assigned when $C_0+C_1+...+C_{k}\neq0$. – user119256 Jan 11 '14 at 11:29
• @AlexRavsky Thanks! I've emphasized that I meant "given the $a$, can one construct two summation methods?" This likely has a similar solution, however. – Sharkos Jan 11 '14 at 17:36