1
$\begingroup$

I am not well versed in summation methods or complex analysis, so I will be presenting a detailed view of my question with examples to illustrate my point as well as a few guiding questions that got me to this point. The guiding questions are mostly wishful thinking on my part. I apologize in advance for the lack of focus on a specific question in the post, but I would prefer responses address the question given by the title if the answer is known, and comment on the line of guiding questions if the answer is unknown.

One specific way of assigning a value to a series that is divergent in the usual sense of partial sums is to see that it matches an expression for another series given in terms of some parameter or variable. As an example, consider the following series. $$1-1+1-1+...$$ This series could be identified as the following series evaluated at $z=-1$. $$\sum\limits_{k=0}^\infty {z^k}$$ This series is the well known geometric series, and is analytic on some open set of the complex plane, but -1 is not part of that set. As such, we analytically continue the function, which provides us with the function $f(z) = \frac{1}{1-z}$. This is useful because this function is well defined at the point $z=-1$ of interest, and we therefore assign the value $f(-1)=\frac{1}{2}$ to our original series.

What seemed important here were the following facts:
1.) We found a function (in this case a series) with series representation that matches our series of interest, term-by-term for the value of some input.
2.) The function we found could be analytically continued to a region of the complex plane with a well-defined value at our input of interest.

I am moving forward with the assumption that these are the only two requirements for the general technique that I refer to as "analytic continuation as a summation method". This led to the following line of questioning which inspired the title of this post.

Given a series $S$, is the existence of a function satisfying properties 1 and 2 guaranteed? If so, is the function unique? If not, is the value obtained for all such functions the same so that this method could be considered well-defined? If this is not true for all series, is there a known subset $U$ for which is is true?

Related but somewhat tangent questions that probably deserve their own post but which feel contingent on this post would include: For each absolutely convergent series in $U$, does the value obtained by analytic continuation agree with the standard sum? For two conditionally convergent series $S_1$ and $S_2$ in $U$ that are permutations of each other, does the analytic continuation method always assign the same value? I ask this question in particular because the standard sum would not satisfy this property in general due to the Riemann rearrangement theorem.

I've failed to find a counterexample that satisfies both properties, but there is an interesting series which has a function satisfying properties 1 and 2 but also has another that only satisfies property 1. $$1+1+1+1+...$$ The above series could be attributed to a geometric series with ratio 1, but this in turn is still a point where the analytic continuation of the geometric series fails to give a finite value. However, if we use the series $$\sum\limits_{n=1}^\infty \frac{1}{n^s}$$ we recognize that when $s=0$, we get term-by-term matching. So, we analytically continue this function to obtain the Riemann zeta Function, which gives $\zeta (0)=-\frac{1}{2}$. It would be interesting if there was a well-motivated way to obtain the value $-\frac{1}{2}$ from the geometric series continuation that worked for other such functions that satisfy property 1 but not property 2, but that is just some more wishful thinking beyond the already wishful thinking of my previous line of reasoning.

$\endgroup$
  • $\begingroup$ This has nothing to do with sums. Any time you have a function defined on a region of points, you can analytically continue it. The answer to whether properties 1 and 2 can always be guaranteed is "no". You can have a series converge only at one point, for instance, and then there is no way to analytically continue it uniquely. To analytically continue a function uniquely, you need to know its values on an open set generally $\endgroup$ – mathworker21 Nov 3 '18 at 19:18
  • $\begingroup$ You have highlighted why a function that satisfies 1 might not satisfy 2, but you don't seem to have provided reasoning as to whether there exists a function that satisfies both. $\endgroup$ – cpoole Nov 3 '18 at 19:26
  • $\begingroup$ Look at mathworld.wolfram.com/NaturalBoundary.html $\endgroup$ – saulspatz Nov 3 '18 at 19:42
  • $\begingroup$ @cpoole you do not understand my comment. $\endgroup$ – mathworker21 Nov 3 '18 at 19:54
  • 1
    $\begingroup$ That comment constitutes an acceptable counterexample, so feel free to post it as a fully fledged answer and I'll accept it. I agree that limiting the method to Dirichlet series and power series is interesting and encourage you to discuss it if you post an answer, but it wouldn't be necessary for the answer to be accepted. $\endgroup$ – cpoole Nov 3 '18 at 21:23
0
$\begingroup$

Here's the classical counterexample to your question. It was an important topic of discussion in the 18th century and the early 19th century, long before any infinite series had a firm foundation, let alone divergent series. It's adapted from Wikipedia:History of Grandi's series. Consider your motivating series: $$1-1+1-1+1-1+\cdots$$ Define the functions $f$ and $g$ on the open unit disk $|z|<1$: $$f(z)=1-z+z^2-z^3+z^4+z^5+\cdots=\frac{1}{1+z}$$ $$g(z)=1-z^2+z^3-z^5+z^6-z^8+\cdots=\frac{1+z}{1+z+z^2}$$ We don't have to analytically continue very far! Just take a limit at the boundary of the original domains of $f$ and $g$. Sadly, we get $\lim_{z\to1}f(z)=\frac12$ versus $\lim_{z\to1}g(z)=\frac23$.

To paraphrase Lagrange, the problem is that the series used to describe $g$ is not actually a power series. So this example shows why you'll need to restrict the kinds of series under consideration in order to get an interesting result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.