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The following words reflect my understanding(an elementary one) of the divergent series. We first define an infinite series as follows:

$L = \sum_{n=0}^{\infty}a_n \Leftrightarrow L = \lim_{k \rightarrow \infty} S_k.$

Where $S_k$ is the partial sum of the infinite series from $a_0$ to $a_k$. A series whose limit exists is said to be convergent, if not, then it's called divergent.

By this former definition, series like:

$1-1+1-...$ and $1+2+3+...$ are divergent.

Then we have the notion of regularized sum. Where we look for a new definition for infinite series such that it allows us to assign real values to some divergent series. Also in the new definition series that are normally convergent under the definition $L = \sum_{n=0}^{\infty}a_n \Leftrightarrow L = \lim_{k \rightarrow \infty} S_k$, are convergent under the new definition, and the two definitions yield the same exact limit $L$ for the normally convergent series. Although I'm not sure of following, but different summation methods always assign the same value for a divergent series(in case it can be assigned to), so that $1-1+1-...=1/2$ under Caesaro summation and Abel's and any other summation that assign a value to such series.

In addition to that, there are series like $1+2+3+...$ , that are not Caesaro or Abel summable, but can be summed under other methods like zeta regularization; This implies that a series that is not summable under certain summation method(say Caesaro's), can be summable under other summation methods(like zeta).

This last fact leads me to my question:

-Can every divergent series be regularized? That is, for every series that is not summable under certain summation methods, can we find a new summation method that sums it up?

-If the answer is yes to the last question, then, does there exist a summation method such that it can sum(regularize) every single divergent series?

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In the most general sense, a summation is a partial function from the set of summand sequences to $\mathbb R$ (or $\mathbb C$). This sounds like we could assign more or less arbitrary values and if we want we really can. However, certain properties of summations are preferred to hold, such as

  • Regularity that is, our summation method should be an extension of the standard -convergent-sequence-of-partial-sums method
  • Linearity that is, if we define $\sum a_n$ and $\sum b_n$ then we also define $\sum(ca_n+b_n)$ and have $\sum(ca_n+b_n)=c\sum a_n+\sum b_n$
  • Stability $\sum a_n$ is defined if and only if $\sum a_{n+1}$ is defined and we have $\sum a_n=a_2+\sum a_{n+1}$

To repeat: not all summation methods (not even all methods in practical use) obaey all three criteria. But if we concentrate on methods obeying all three then indeed we often get that certain (classically) divergent series are always assigned the same value under any summation method. For example, $\sum x^n=\frac1{1-x}$ follows for all $x\ne 1$ where we define the sum by, merely playing around with stability and linearity.

So how high can we try? We can use Zorn's lemma to find a maximal regular, linear, stable summation method. But will "maximal" imply "total", i.e., that all series become summable? And will the summation thus obtained be well-defined? Unfortunately, the answer to both is no. This can already be exemplified with $\sum 1$, which has do be a solution of $\sum 1 = 1+\sum 1$ per statbility. (Then again, you have have read that regularization can assign $1+1+1+\ldots =-\frac12$; apparently those methods are not linear or nopt stable ...)

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  • $\begingroup$ First of all, thank you for your enlightening answer. In the stability condition, shouldn't $\sum a_n=a_0+\sum a_{n+1}$? is $a_2$ a typo? You ask a series being summable on one hand, and its summation being well-defined on the other hand in two separate questions(as being different notions); The two questions are equivalent, right? I don't understand what you mean by the last example of $\sum 1$, can you explain? $\endgroup$
    – Omar Nagib
    Sep 27, 2015 at 20:39
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    $\begingroup$ What happens if we get rid of the stability hypothesis ? Can we get a total function? $\endgroup$
    – AlienRem
    Jan 9, 2017 at 0:16
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A new method for summing series is available with the hyperreal numbers, but depends on an assumption that the limit of $-1^\omega = 0$, where $\omega$ is the hyperreal unit infinity. Essentially, the steps are:

  1. Take the series representation, and derive a formula for the sum at the point of the $k$th term. This is done by using discrete calculus, or a standard summation method.
  2. Substitute the infinite unit $\omega$ for $k$
  3. Reduce the result using appropriate rules, including $-1^\omega = 0$.
  4. Optionally, round to the whichever "level of infinity" you desire (i.e., take the standard part, or leave it with infinities/infinitesimals, etc.

Example:

Let's say you wanted to know the sum of the natural numbers: $1 + 2 +3 + \ldots$. This is a simple arithmetic series. The formula for the $k$th partial sum of an arithmetic series is $$\sum_{n = 1}^k a + (n - 1)d = \frac{k^2d}{2} + \frac{k(2a - d)}{2}$$

So, since we are going to infinity (i.e., $\omega$), and $a = 1$ and $d = 1$, then the result will be $\frac{\omega^2}{2} +\frac{\omega}{2} \simeq \frac{\omega^2}{2}$ (the $\simeq$ indicates that the principal value of the result is $\frac{\omega^2}{2}$ - once you square an infinity, the next infinite power down plays essentially no role in the value).

Now, let's look at the series $1 + 3 + 5 + \ldots$. Here, $a = 1$ and $d = 2$, so the result will be simply $\omega^2$.

Using this method, you can do things like divide series by each other. So, we can say: $$\frac{(1 + 2 + 3 + \ldots)}{(1 + 3 + 5 + \ldots)} = \frac{\frac{\omega^2}{2}}{\omega^2} = \frac{1}{2}$$

Using this method, you have a lot of better-behaved divergent series. In this method, $(1 + 3 + 5 + \ldots)$ is not the same series as $(1 + 0 + 3 + 0 + 5 + \ldots)$, even though they look like they might be. The latter has a value of $\frac{\omega^2}{4}$.

This method winds up yielding consistent values over a wide range of divergent series.

You can see the introduction to the method, and an application of it to the question of Ramanujan summation here.

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  • $\begingroup$ The standard value of $1+2+3+... = -\frac{1}{12}$ and the same family of techniques assign $1+3+5+... = \frac{1}{3}$ it seems the ratio between these is $-\frac{1}{4}$ which yields a different interpretation than what you have here. So while I think this has a lot of merit (its a very natural generalization of limits and singularities) I don't think its consistent with results from complex analysis and ramanujan summation $\endgroup$ Jul 26, 2023 at 0:06
  • $\begingroup$ @SidharthGhoshal - Ramanujan summation is a normalization, not a sum in the traditional sense, and therefore would not work with this. For an analysis of the question of the sequence 1 + 2 + 3 + ..., you should see this paper of mine: academia.edu/37974912/Numberphile_s_Proof_for_the_Sum_1_2_3_ $\endgroup$
    – johnnyb
    Jul 26, 2023 at 14:09
  • $\begingroup$ In your paper you mention "However, the question still remains why physicists can use −1/12 as a stand-in for the sum of all natural numbers... While no conclusive reason for this has been established" (correct me if i'm wrong) but I believe we have VERY conclusive reasons why it works. As you noted in the next section the cosine smoothed sum of $1+2+3...$ has an asymptotic expansion given by $T(n)-1/12 + O(1/n)$ where $T$ is some growing function. In situations like the Casimir Effect calculation I believe the quantity being evaluated implicitly is $1+2+3... -B(n)$ where $B(n) \sim T(n)$ $\endgroup$ Jul 26, 2023 at 15:44
  • $\begingroup$ And so the quantity being measured ends up being the difference of the constant level asymptotic parts of both series which ends up as $-\frac{1}{12} - 0$. Terence tao's blog goes into this a bit: terrytao.wordpress.com/2010/04/10/… $\endgroup$ Jul 26, 2023 at 15:45
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    $\begingroup$ Also $(-1)^\infty = 0$ follows from the fact that $\sum_{n=0}^{\infty} x^n = \frac{1-x^{\infty+1}}{1-x} = \frac{1}{1-x}$ see this post here: mathoverflow.net/questions/429326/… where the fact $e^{e^\infty} = 1$ is discussed. In general $e^{e^{\infty}} = e^{0} = 1$. I noticed this was ongoing work in your paper incase it hasn't already been resolved $\endgroup$ Jul 26, 2023 at 15:57
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We hope so in general, but as of today we don't have techniques for many divergent series. Consider something like $ \sum_{n=0}^{\infty} 2^{2^n}$. Having an agreed upon regularization for this would amount to having defined the function $\sum_{n=0}^{\infty} z^{2^n}$ beyond its natural boundary and unfortunately there is no such standard agreed upon continuation to my knowledge.

In general while many regularization techniques can be put forth there is a "naturality" to some regularizations. In the case of the famous $1+2+3+4+... = -\frac{1}{12}$ for it NOT to "formally" equal $-\frac{1}{12}$ would make the laurent expansion of $\frac{1}{1-e^x}$ difficult to explain because the linear term there is $-\frac{1}{12}$ and there formal manipulations that equate it with $1+2+3+...$ see here on why this should be the case.

The reason I bring up naturality is we don't have a "natural" continuation of that earlier power series to my knowledge so at this time we don't have a natural way to regularize the divergent series that come from it.

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To be overly pedantic, the summation method $S_0$ defined as follows $$S_0 \left( \sum a_n \right) = \begin{cases} \sum a_n & \text{if it converges} \\ 0 & \text{otherwise} \end{cases} $$ sums every divergent series. Of course, $S_0$ is quite useless. But why is it wrong? To actually show that $S_0$ isn't a good divergent series summation, we should find a different divergent series summation method that we trust, and check whether $S_0$ agrees with that method.

There is one summation method that we generally trust. Assume we are also looking at a function, rather than a specific point. Then
$$S_1 \left( \sum a_n \right) = \begin{cases} \sum a_n & \text{if it converges} \\ \text{Analytical continuation of } \sum a_n & \text{if it exists} \end{cases}$$ For example, if we are looking at the function $\sum n^s$, then $S_1(\sum n) = -\frac{1}{12}$, using the analytical continuation of the zeta function. Many divergent series methods you will run into are basically attempts from people to create something that behaves like $S_1$. However, most people don't realize that essentially every divergent series summation you run into ends up being incomparable to $S_1$. I.e. there tends to exist sums that the divergent series method can sum that cannot be obtained through analytical continuation. For example, consider the (very weak) summation method, Cesaro summation. Take $f(x) = \sum_{n=1}^\infty (-1)^n \frac{x^n}{1-x^n}$

$f(x)$ cannot be analytically continued to $|x|>1$, since it has a natural boundary due to the dense poles coming from $1-x^n = 0$. However, for sufficently large $n$, this sum looks just like $\sum (-1)^n$, which can be summed using Cesaro summation. Generally, if we a divergent series summation method that can sum $\sum a_n$, then (with some reasonable conditions on the properties of the summation method) $\sum a_n \frac{x^n}{1-x^n}$ is a function that cannot be analytically continued, but can be summed using divergent series summation.

If you look at the problem more closely, you'll see that even regular summation is incomparable to analytical continuation. For example $\sum e^{-x^n}$ has a natural boundary at $|x|=1$, but of course the sum still converges beyond this.

So, the problem isn't so much that we can't regularize divergent series, but that there isn't a framework to understand the right regularization for divergent series. When we restrict ourselves to regularizing series that have an associated analytical continuation, then we have the whole theory about consistency of analytical continuation, and various knowledge about how they interact with differential equations, composition, etc. We have specific knowledge on the type of coherence we obtain from analytical continution, but there still lots of work to do to understand to coherence we would achieve from continuations by divergent series summation.

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