Talk:Radius of convergence

L and C: root and ratio test
Why must 1/L and 1/C be used for the radius of convergence? Why can't we invert the equations as they stand? For example $$ {1 \over L} = R = \mbox{Radius Of Convergence} = \lim_{n\rightarrow\infty} \left| \frac{c_{n}}{c_n+1} \right| $$. Why can't we put that up. Do C and L have special uses?? - Fresheneesz


 * Alright that it, I'm freaking changing it. If someone wants to change it back, discuss it *here*. Fresheneesz 08:25, 27 March 2006 (UTC)


 * A possible reason is consistency with root test and ratio test. But feel free to change it. -- Jitse Niesen (talk) 12:03, 27 March 2006 (UTC)


 * Ratio test doesn't have a letter assocaited with it on the page at least (anymore?). But I don't want to change it if people actually use the letters L or C to mean something. If its 100% arbitrarily used.... Well, yea i guess it would be a good parallel. I think i'll fix it up so it is less arbitrary. Fresheneesz 09:23, 28 March 2006 (UTC)

Radius of convergence of other series
I've looked it up, and I can't seem to find any reference to the radius of convergence for series that are not power series. Is radius of convergence only definined for power series? Also, the ratio and root tests given on this page are not the same as the ratio and root tests that this page links to (I think), because here, the root test is applied to only the coeffecients of the term of the power series, while on the root test page, the entire term is applied. Fresheneesz 01:21, 29 March 2006 (UTC)


 * Well, you have to do a bit of work. A power series is a series of the form
 * $$ \sum_{n=0}^\infty c_n z^n. $$
 * I set the centre a to zero because it does not matter. As you say, the root test should be applied to the whole term, so the root test says that the series converges if
 * $$ \limsup_{n\to\infty} \sqrt[k]{|c_k z^k|} < 1. $$
 * This condition is equivalent to
 * $$ |z| < \frac{1}{\limsup_{n\to\infty} \sqrt[k]{|c_k|}}, $$
 * which is the formulation on this page.
 * For a general series depending on a variable z, the region of convergence will most likely not be a circle, and thus it makes no sense to talk of a radius of convergence. -- Jitse Niesen (talk) 07:29, 29 March 2006 (UTC)


 * Oh, thats interesting, I should have seen that. That would be an insightful addition to this page, I think i'll add it. I noticed that "region of convergence" redirects here, while it has no explanation whatsoever. I don't know what I can about it, but something should be added about it. Fresheneesz 10:19, 29 March 2006 (UTC)

Merge with power series
Since it looks as if a "radius of convergence" is a term only used for power series, and that page already has something on radius of convergence, it seems only natural to merge the two. Any comments? Fresheneesz 10:29, 29 March 2006 (UTC)


 * I disagree. This is an important enough concept to warrant its own article. Besides, the radius of convergence is only for power series of one variable, while power series can be in many variables. Oleg Alexandrov (talk) 17:42, 29 March 2006 (UTC)
 * I bet in power series of many variables one has many radii to define a polydisc of convergence, or even polyballs("Hartogs stuff') or more general cartesian products of balls of different dimension.Rich 21:33, 31 October 2006 (UTC)

odd wording
"The root test is defined as: C=..." - That doesn't make sense. A test is not an equation, nor is it a real number. Also for "ratio test" later. --Zerotalk 08:59, 14 August 2006 (UTC)


 * I agree. I have no idea what is being said. Can someone fix this? A5 21:06, 31 October 2006 (UTC)

I also agree. I've done some cleanup accordingly. The article titled root test actually asserted that the root test is a number. I've done some editing on that one too. Michael Hardy 21:54, 31 October 2006 (UTC)

3-sphere of convergence?
Suppose one had a power series w/radius of conv = 1, say, around z=0, say. If z were allowed to be quaternionic, then the boundary of convergence would be a 3 sphere of rad 1. I bet the subset of the boundary where the series converges could be quite beautiful and fascinating. Does anyone know of results/research in this area? It would be a great addition to put in this article.Rich 07:47, 25 September 2006 (UTC)

What about
What about $$ |z-a| =r \!$$? --Abdull 17:38, 11 February 2007 (UTC)


 * That's a limiting case. Many things can happen depending on the function. You may have divergence, or convergence, or conditional convergence. Oleg Alexandrov (talk) 00:07, 12 February 2007 (UTC)


 * In fact, the series MUST be bad at at least one point on the circle |z-a|=r since the radius of convergence is equal to the distance from point "a" to the nearest singularity. --Rocketman768 18:52, 1 May 2007 (UTC)
 * I think you mean "to the nearest SINGULAR POINT", not "singularity". It sounds like you're saying by "series MUST be bad at at least one point" that the series must diverge somewhere on the circle of convergence, which I disagree with, see article. Also see Konrad Knopp, Theory of Functions Part I, Dover, page 100, and Exercise 2 on page 73. --Rich Peterson

clarity and simplicity section is wrong, i think
many series have no point on circle of conv where it blows up. But that is what i think the section is saying. I put expert needed tag on it.Rich (talk) 03:40, 24 June 2008 (UTC)


 * It seems right to me. Where do you think the section says that all series have a point on the circle of convergence where they blow up? -- Jitse Niesen (talk) 13:19, 24 June 2008 (UTC)

It doesn't say "blows up"; it says "has a non-removable singularity". Rich, can you cite an example of a power series with a finite radius of convergence that has no non-removable singularity on the boundary? Michael Hardy (talk) 14:47, 24 June 2008 (UTC)
 * You must realize that I'm at the outer limits of my knowledge here. Part of why I think it says they must blow up(that is, diverge) somewhere on the circle is that both examples that are given do so. My mis?understanding of the situation is that there will always be at least one "singular point" (not SINGULARITY)on the circle of convergence, even though the series may converge everywhere on the circle. "Singular point" doesn't mean a point where the function is undefined, rather I think it means a point from which analytic continuation is impossible. (See Knopp, I'll try to get the page number asap.) For an example where I see no non-removable singularity(which I interpret as failing to converge, that is, "blowing up"), consider ([z^n]/[n^2]), summed from n=1 to infinity.

Even if I'm wrong about this, hopefully I'll learn something. RegardsRich (talk) 19:13, 25 June 2008 (UTC) "Exercise. The unit circle is the circle of convergence of the power series [sum z^n/n^2 for n=1 to infinity].  Show that the point +1 is a singular point of the function represented by the series in the unit circle, by expanding in a new power series with center z1 = +1/2. (Nevertheless, the given series is convergent for z = +1!!)"Rich (talk) 20:06, 25 June 2008 (UTC)
 * Theory of Functions by Konrad Knopp, Part I, translated by Frederick Bagemihl. 1945, Dover. Pages 99-104 are relevant. On page 104:


 * Well, you quote Knopp as saying that 1 is a singular point, and if Knopp is right, then that is not an example that shows the article is wrong, but is in perfect agreement with the article. The article says there must be a singular point on the boundary.  In this case, the center is 0 and the radius of convergence is 1, so the number 1 is indeed on the boundary, and (according to Knopp) the number 1 is a singular point.  So there you have a singular point on the boundary.  In order to show that the article's statement is wrong, you'd need to give an example in which there is NO singular point on the boundary. Michael Hardy (talk) 20:17, 26 June 2008 (UTC)
 * Perhaps you're right, but I think attention should be drawn to the distinction between a singular point and a singularity. Singularity is linked in the clarity & simplicity section, but I don't see any example in the singularity article that resembles Knopp's "singular point." Perhaps the clarity&simplicty sect. should have an example like Knopp's above. Apart from that, I think scholarly method and humility support keeping the experts needed tag. What do you think?Regards, Rich (talk) 00:43, 27 June 2008 (UTC)
 * In addition, maybe we should start a new article on "singular point" in Knopp's sense, or put it in a section of analytic continuation.Rich (talk) 00:51, 27 June 2008 (UTC)

I'm not sure what you mean by this distinction. Maybe I'll take a look at Knopp and see what he says. I confess I've never seen an example quite like this. At any rate, this function cannot be extended to a holomorphic function defined on any disk centered at 0 whose radius is more than 1, and not merely because of a failure of the series to converge. Michael Hardy (talk) 02:51, 27 June 2008 (UTC)


 * [sum z^n/n^2 for n=1 to infinity] defines a function called the dilogarithm. This function has a branch point at z = 1, which is one of the possibilities described at Mathematical singularity. This article has a very similar example, example #3 at Radius of convergence. Perhaps it would help if these three examples include an explanation of the singularities involved? -- Jitse Niesen (talk) 12:41, 27 June 2008 (UTC)
 * I see that the article you linked to doesn't mention the branch point. Michael Hardy (talk) 16:45, 27 June 2008 (UTC)

"... However, we could also have found the radius of convergence by noting that the function $$f(z)$$ has singularities at $$\pm i$$, which are at a distance 1 from 0."
 * The current version, on 7/15/2008, has:

This makes it clear the radius of convergence must be less than or equal to 1. But to play devil's advocate, why must it equal 1? Couldn't there be singular points, perhaps of the type mentioned above by Konrad Knopp, still closer to the origin? Thanks.Rich (talk) 18:45, 15 July 2008 (UTC)

A vocabulary point -- there are three types of possible singularities in the complex plane. One of the glories of complex analysis is that there are no other singularities. Within the radius of convergence of a power series (not on the boundary, but the interior of the circle), the series converges absolutely, hence it is bounded. Thus, the only possible singularities are of the removable type, and they can be removed, leaving a holomoprhic function behind. Best, RayAYang (talk) 17:43, 17 July 2008 (UTC)
 * 1) Removable singularities. These are equivalent (by a theorem whose proof I do not have at hand) to functions which are bounded in a neighborhood of the singularity. You can "remove" the singularity by defining the function to be its limit at that point; there will be a unique limit - this is a theorem by Riemann.
 * 2) Poles. The function will "blow up" to infinity along every curve of approach to the singularity.
 * 3) Essential Singularities. The function will, in an arbitrarily small neighborhood of the singularity, assume every possible value except one. Infinitely often. This is a highly nontrivial result by Picard. An example of such a function is $$e^{1/z}$$ about 0 (try approaching 0 along the real line from either direction. Then do the same along the imaginary directions -- you will see that in some directions there is an limit as you approach, in other directions it diverges, but along the imaginary axis it is oscillating but of bounded modulus).
 * That may be all of the "singularities" but it's not all of the singular (nonremovable)points.Regards,130.86.14.87 (talk) 03:41, 18 July 2008 (UTC)Rich (talk) 03:42, 18 July 2008 (UTC)
 * No, it is all of the possible singularities. Essential singularities are technically defined as all singularities that are neither poles nor removable, and then their other properties are proved from that definition. In the example given above, since the power series has a radius of convergence of 1 about 0, that means there can be no non-removable singularities inside the disc of radius one centered at 0. I don't know how to phrase it more clearly. Can you try rephrasing and explaining your question at length? Best, RayAYang (talk) 04:02, 18 July 2008 (UTC)

Many apologies to all concerned. I copied over the text of the "theorem" stated in the old version of the article without thinking it through. There is no requirement that there be a non-removable singularity on boundary of the circle of convergence, merely that a point exists where the function can no longer be defined holomorphically. This situation may arise in the case of functions with branch cuts, such as the square root function or the logarithm defined over the complex numbers.

With hopes of ending the confusion over what's been happening, I will give a more formal summary. There are two theorems that are relevant here: From these two theorems we conclude that at least one point on the boundary of a power series' domain of convergence (which is a disc), must be on the boundary of the set where the function can be holomorphically defined. This need not mean a singularity -- there are ways to define holomorphic functions on domains that leave out a good deal more than a point. I hope this clears up the confusion over this article. Best, RayAYang (talk) 04:17, 18 July 2008 (UTC) RayAYang (talk) 04:17, 18 July 2008 (UTC)
 * 1) Power series define a holomorphic function within their domain of convergence, which is always a disk of some radius (possibly null, possibly infinite). Anything can happen on the boundary of the disk -- the series can converge everywhere, diverge everywhere, converge at some points but not others, etc.
 * 2) Within any disc (as large as you wish) strictly contained within a domain where a function is holomorphic, the power series representation of the function, taken about the point at the center of the domain, is convergent.


 * You wrote:
 * This situation may arise in the case of functions with branch cuts, such as the square root function or the logarithm defined over the complex numbers.
 * Does that mean you don't consider a branch point to be a non-removable singularity?
 * This need not mean a singularity -- there are ways to define holomorphic functions on domains that leave out a good deal more than a point.
 * Can you cite examples of the sort of thing you have in mind? Michael Hardy (talk) 04:27, 18 July 2008 (UTC)
 * The complex square root is an excellent example, I think. A pretty standard way to define it is to define it everywhere except along the negative part of the real axis. This example at MathWorld provides such a case. In this case, if we were to take a power series centered about 1, it would have a radius of convergence 1, the series would converge properly to 0 at 0, but we couldn't extend it further. 0 is not a singularity of the square root so defined -- we converge to 0 along every possible curve approaching it. However, you cannot produce a power series of positive radius of convergence, for this definition of the complex square root, in any open neighborhood of 0. RayAYang (talk) 04:34, 18 July 2008 (UTC)
 * This need not mean a singularity -- there are ways to define holomorphic functions on domains that leave out a good deal more than a point.
 * Can you cite examples of the sort of thing you have in mind? Michael Hardy (talk) 04:27, 18 July 2008 (UTC)
 * The complex square root is an excellent example, I think. A pretty standard way to define it is to define it everywhere except along the negative part of the real axis. This example at MathWorld provides such a case. In this case, if we were to take a power series centered about 1, it would have a radius of convergence 1, the series would converge properly to 0 at 0, but we couldn't extend it further. 0 is not a singularity of the square root so defined -- we converge to 0 along every possible curve approaching it. However, you cannot produce a power series of positive radius of convergence, for this definition of the complex square root, in any open neighborhood of 0. RayAYang (talk) 04:34, 18 July 2008 (UTC)
 * The complex square root is an excellent example, I think. A pretty standard way to define it is to define it everywhere except along the negative part of the real axis. This example at MathWorld provides such a case. In this case, if we were to take a power series centered about 1, it would have a radius of convergence 1, the series would converge properly to 0 at 0, but we couldn't extend it further. 0 is not a singularity of the square root so defined -- we converge to 0 along every possible curve approaching it. However, you cannot produce a power series of positive radius of convergence, for this definition of the complex square root, in any open neighborhood of 0. RayAYang (talk) 04:34, 18 July 2008 (UTC)

Hah. Changed the article again -- it's not that the function can't be defined at a point on the circle, it's that the function cannot be defined past the circle. Ugh. My apologies to everybody on this project. I hope we'll get there eventually -- please do keep checking my work :) RayAYang (talk) 04:43, 18 July 2008 (UTC)

Example 4 in Convergence on the Boundary is a series of functions, but--
But I think it's not a power series, since it's not absolutely convergent. If you rearrage the terms so that it's a power series, you risk losing convergence on the boundary or at least changing what it converges to, thus it's a different function than the power series you rearranged it to. Shouldn't this example be labelled as a more general object than power series? Rich (talk) 08:28, 16 March 2010 (UTC)


 * Absolute convergence is not a necessary condition for being a power series; it is a theorem. The terms are already arranged so that the series is a power series. It's written with several powers of z in each summand because this way of writing it avoids some complications with base 2 logs and floor functions. (Also, that's the way the original source has it.) Ozob (talk) 11:02, 16 March 2010 (UTC)
 * Lots of power series don't converge, so I think you are saying by "it is a theorem" that there is a theorem if a power series does converge at a point, then it converges absolutely. However, in example 4 it says it doesn't converge absolutely on the boundary. If this is true, and if the theorem which I am supposing you are citing is correct, then Example 4 is not a power series on the points of the boundary. But it may be that you meant something slightly different was a theorem. Best wishes, Rich (talk) 19:37, 16 March 2010 (UTC)

The theorem is that if a power series converges at a point in the interior of the circle of convergence, then it converges absolutely at that point. That leaves open the question of a power series converges absolutely at boundary points where it converges. A simple example is
 * $$ \sum_{n=1}^\infty \frac{z^n}{n}. $$

This converges absolutely for |z| < 1. It diverges at z = 1. It converges conditionally at z = &minus;1. Michael Hardy (talk) 20:04, 16 March 2010 (UTC)
 * Right, I was just about to post an addendum to that effect as it had finally occurred to me that if this is the theorem Ozub was referring to, it was not meant to hold on the boundary. But then Example 4 is quite possibly not equivalent to the power series one would obtain if one rearranged and collected its terms its so as to have exponents in increasing order. That may be why the authors in the original source wrote it like this--it was constructed to show that a series of functions could converge uniformly but not absolutely on the boundary--and they had been unable(it may even be impossible, afik)to find a power series with that property.Rich (talk) 20:52, 16 March 2010 (UTC)


 * Yes, absolute convergence in the interior is the theorem I'm thinking of. I think the confusion is that example 4 is a power series with its exponents in increasing order, but it's not written in the usual way because the usual way is more complicated. Let me demonstrate:
 * $$\begin{align}

P(z)&=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{2^n\cdot n}(z^{2^{n-1}} + z^{2^{n-1}+1} + \cdots + z^{2^n-1})\\ &= \sum_{n=1}^\infty \frac{(-1)^{n-1}}{2^n\cdot n}(z^{2^{n-1}} + z^{2^{n-1}+1} + \cdots + z^{2^{n-1} + 2^{n-2} + \cdots + 2^1 + 1})\\ &= \sum_{n=1}^\infty \frac{(-1)^{\log_2 2^{n-1}}}{2^n\cdot(\log_2 2^{n-1} + 1)}z^{2^{n-1}} + \frac{(-1)^{\log_2 2^{n-1}}}{2^n\cdot(\log_2 2^{n-1} + 1)}z^{2^{n-1}+1} + \cdots + \frac{(-1)^{\log_2 2^{n-1}}}{2^n\cdot(\log_2 2^{n-1} + 1)}z^{2^{n-1} + 2^{n-1} \cdots + 2^1 + 1}\\ &= \sum_{n=1}^\infty \frac{(-1)^{\lfloor\log_2 2^{n-1}\rfloor}}{2^n\cdot(\lfloor\log_2 2^{n-1}\rfloor + 1)}z^{2^{n-1}} + \frac{(-1)^{\lfloor\log_2 (2^{n-1} + 1)\rfloor}}{2^n\cdot(\lfloor\log_2 (2^{n-1} + 1)\rfloor + 1)}z^{2^{n-1}+1} + \cdots + \frac{(-1)^{\lfloor\log_2 (2^{n-1} + 2^{n-2} + \cdots + 2^1 + 1)\rfloor}}{2^n\cdot(\lfloor\log_2 (2^n + 2^{n-1} + \cdots + 2^1 + 1)\rfloor + 1)}z^{2^{n-1} + 2^{n-1} \cdots + 2^1 + 1}\\ &= \sum_{n=1}^\infty \frac{(-1)^{\lfloor\log_2 2^{n-1}\rfloor}}{2^{\lfloor\log_2 2^{n-1}\rfloor + 1}\cdot(\lfloor\log_2 2^{n-1}\rfloor + 1)}z^{2^{n-1}} + \frac{(-1)^{\lfloor\log_2 (2^{n-1} + 1)\rfloor}}{2^{\lfloor\log_2 (2^{n-1} + 1)\rfloor + 1}\cdot(\lfloor\log_2 (2^{n-1} + 1)\rfloor + 1)}z^{2^{n-1}+1} + \cdots + \frac{(-1)^{\lfloor\log_2 (2^{n-1} + 2^{n-2} + \cdots + 2^1 + 1)\rfloor}}{2^{\lfloor\log_2 (2^{n-1} + 2^{n-2} + \cdots + 2^1 + 1)\rfloor + 1}\cdot(\lfloor\log_2 (2^n + 2^{n-1} + \cdots + 2^1 + 1)\rfloor + 1)}z^{2^{n-1} + 2^{n-1} \cdots + 2^1 + 1}\\ &= \sum_{m=1}^\infty \frac{(-1)^{\lfloor\log_2 m\rfloor}}{2^{\lfloor\log_2 m\rfloor + 1}\cdot(\lfloor\log_2 m\rfloor + 1)}z^m \end{align}$$
 * From this it's pretty straightforward to see that the radius of convergence is one. But in order to really show the facts about the boundary, I think you want it in the form given in the article. Ozob (talk) 23:02, 16 March 2010 (UTC)
 * That's pretty nice mathematics! I'm sure you're correct that the radius of convergence is 1, and I agree that we want it in the form given in the article, and I think it should be retained in the article. But, because it a series of of polynomials rather than a series of increasing powers of z, it's already "nominally" not a power series. When you add in that it is conditionally convergent on the boundary, you have a series that probably can't even be rearranged to be a power series without changing the values of the function on boundary points. So I think it should be pointed out that it isn't a power series, though closely related. Best wishes, Rich (talk) 22:18, 17 March 2010 (UTC)
 * OK, I still disagree with you on this one. I agree that it's written as a series of polynomials. I also agree that the given expression is not the usual expression for a power series. But I still believe that the series is a power series; it's just had some of its consecutive terms collected. My point is that no term rearrangement is necessary: The n&thinsp;th term of the series contains only powers of z that are at least 2n &minus; 1, and these terms are already in order. A perhaps more convincing reason is that you can represent the series as a sequence of partial sums Sp. If you call the partial sums of the power series at the bottom of the above derivation Tq&thinsp;, then we have $$S_p = T_{2^p-1}$$. So these two sequences of partial sums have the same limit, and hence their behavior at the boundary is the same.


 * I may as well include the justification for the radius of convergence being one: Use the ratio test. The ratio has a factor of the form 2 to the power $$\lfloor\log_2 m\rfloor - \lfloor\log_2 (m + 1)\rfloor$$. For all m sufficiently large, this quantity is either zero or minus one because $$\log_2 m/(m + 1)$$ tends to zero; so 2 raised to this quantity is one or 1/2. The ratio also has a factor of $$(\lfloor\log_2 m\rfloor + 1)/(\lfloor\log_2 (m+1)\rfloor + 1)$$. This tends to one, since eventually either the top and the bottom are either equal or the bottom is larger than the top by one. The ratio has a factor of &minus;1 which we can ignore since we're interested in the absolute value. Finally, we take the lim sup to conclude that the radius of convergence is one. Ozob (talk) 23:29, 17 March 2010 (UTC)

I suggest that in the article the example be displayed like this:
 * $$\sum_{i=1}^\infty a_i z^i \text{ where } a_i = \frac{(-1)^{n-1}}{2^nn}\text{ for }2^{n-1}\le i < 2^n.$$

The difference between this and the current presentation is that it will be more clear that we have to consider convergence of all the partial sums (one for each i) rather than just the convergence of the particular subsequence of partial sums given by the n values. Zerotalk 23:51, 17 March 2010 (UTC)


 * An excellent idea! Done! Ozob (talk) 00:05, 19 March 2010 (UTC)
 * I'm now convinced that you are correct. Also, the suggestion by Zero is good.Rich (talk) 16:26, 21 March 2010 (UTC)

More examples
Can please someone do a few more concrete examples of finding out the radius of convergence for when z is complex and when it is real and a few special cases. It will help a lot. —Preceding unsigned comment added by 128.100.194.208 (talk) 16:07, 4 October 2010 (UTC)


 * What would you like examples of? What sorts of series? (Please keep in mind WP:NOTTEXTBOOK.) Ozob (talk) 22:15, 4 October 2010 (UTC)

Convergence on the boundry.
Can someone actually do the example of showing and actually working out the steps that the power series Σ zn/n converges at every point of the unit circle except at z=1, that the series Σ zn/n2 converges at every point of the unit circle and that Σ nzn does not converges on any point of the unit circle. It is easy to find the radius of convergence but how to determine what happens on the radius of convergence. What is special here is that all of the series mentioned have a radius of convergence of 1. A good idea would also be to include these 3 examples in the relevent section as they are an interesting example due to their 3 properties. —Preceding unsigned comment added by 64.56.250.36 (talk) 07:38, 11 October 2010 (UTC)

Theoretical radius - error?
This section states "The limit involved in the ratio test is usually easier to compute, and when that limit exists, it shows that the radius of convergence is finite." Surely this is incorrect? It's possible that the ratio tends to some <1 value irrespective of the relevant variable, no? 188.74.64.241 (talk) 10:56, 20 March 2016 (UTC)


 * I don't understand your objection, but I believe the article is correct. Ozob (talk) 16:40, 20 March 2016 (UTC)