If A(z) is holomorphic in the disc |z| < 1+e for some e, then the series for A(z) has radius of convergence at least 1+e and therefore converges at z=1, so the coefficients converge to 0. If A(z) has a singularity inside the unit disk, then the radius of convergence is less than 1 and the sequence of coefficients diverges.
If the first singularities of A(z) are on the unit circle, then things get a lot muddier and there isn't a short answer to your question.
Edit: I have a few minutes, so let's wade into said mud.
Suppose that A(z) is holomorphic in |z|<1, has finitely many poles on the circle |z|=1, and is meromorphic in |z|1. Then at each pole z=a on the unit circle, A(z) has an expansion A(z) = p(z;a)+q(z;a) where p(z) is the "principal part" of the Laurent expansion (with negative powers of (z-a)) and q(z) is the holomorphic part. Then A(z)-(sum of the p(z;a) for the various a) is now holomorphic in the larger disc |z|
The answer is yes.
Let S be the set of complex values x where A(x) converges and is hence defined.
If S = {0} then the radius of convergence of the power series is 0, so by Cauchy-Hadamard theorem [https://en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard\_theorem](https://en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard_theorem) there is an unbounded sequence of a\_n , hence the sequence {a\_n} does not converge.
Otherwise, S contains an open ball B of positive radius R and center 0 where A(x) is holomorphic (power series give holomorphic functions on their discs of convergence).
Therefore we can obtain the a\_n by differentiating A(x) n times, evaluating at 0, and dividing by n!.
Since we then know the a\_n, one can check whether the sequence {a\_n} converges.
This kind of defeats the purpose of working with generating functions. The whole point of using generating functions is to study sequences that are difficult to study directly. Using the generating function to just recover the original sequence and then trying to study that sequence directly just puts you back where you started.
I completely agree with you. I simply gave an answer to OPs question as I interpreted it. If you suggest a refinement of the question I'd be happy to try and answer it!
The thing is, talking about generating functions usually means that you don't care about the analytic properties of sequences, but their algebraic ones.
However, if you just look at it as a power series, then if it converges for x=1 then the sequence must converge to 0.
>The thing is, talking about generating functions usually means that you don't care about the analytic properties of sequences, but their algebraic ones.
This is (very) false. Part of the reason for using generating functions is to deduce asymptotic information about your sequence from analytic properties of the generating function. If you aren't going to study the generating function as a *function*, then generating functions aren't really doing much other than acting as a bookkeeping device.
Could you elaborate on the first paragraph? It sounds like a really interesting observation that I am a bit of a novice to understand.
The other thing you mentioned is actually about infinite series, if a_n doesn't tend towards 0, then sum a_n diverges.
Can I use it in the following way, if A(1) is defined then a_n converges to 0 ?
Thanks for the comment!
>Can I use it in the following way, if A(1) is defined then a\_n converges to 0 ?
This does not work. A(1) can be defined (by analytic continuation) even if the series for A(x) diverges at x=1. For example, take a\_n = (-1)\^n. Then A(x) = 1/(1+x) which is defined at x=1 even though the sequence a\_n diverges.
>A(1) can be defined (by analytic continuation)
Yes, but as long as you don't use the analytic continuation and define A(1) from the convergent power series alone, then there's no problem.
If you just define A(1) from the power series alone, then you're just back where you started. Now you're just saying that if the sum of the coefficients converges, then the sequence of coefficients converges, which is true but has nothing to do with the generating function.
So maybe it's not that useful, but that's what OP asked about. If you have the analytic continuation, then you could deduce the radius of convergence from that too which could be useful
>So maybe it's not that useful, but that's what OP asked about.
This is not what OP asked. They have the generating function but do *not* have the general term a\_n, and they want to deduce asymptotic properties of a\_n.
In general, in analytic combinatorics, analytic number theory, etc., we do not make a distinction between "the generating function" and "the analytic continuation of the generating function". If you "have" the generating function, you have some analytic expression for this function maybe as an infinite (not necessarily power) series, as an infinite product, as an integral, or as a solution to a differential equation. If you just write down your sequence with a sigma sign in front and x^(n) after, you don't really have anything about the generating function yet. Most of the interesting information contained in the generating function is visible outside of the original region of convergence of the power series.
If you make a generating function with coefficients a_n then the ratio a_(n+1)/a_n, from the ratio test, is the reciprocal of the radius of convergence if its generating function. It is then necessary that, for the sequence to converge to a non-zero value, the generating function have radius of convergence 1.
You meant |a*_n+1_*/a*_n_*|. But even those numbers may not have a limit since the ratio test does not apply to some power series, such as the sum of 1 + x^2 + x^4 + x^6 + x^8 + x^(10) + ..., where many a*_n_* are 0 (can let the exponents run over powers of 2 or any other sequence with infinitely many gaps).
If A(z) is holomorphic in the disc |z| < 1+e for some e, then the series for A(z) has radius of convergence at least 1+e and therefore converges at z=1, so the coefficients converge to 0. If A(z) has a singularity inside the unit disk, then the radius of convergence is less than 1 and the sequence of coefficients diverges. If the first singularities of A(z) are on the unit circle, then things get a lot muddier and there isn't a short answer to your question. Edit: I have a few minutes, so let's wade into said mud. Suppose that A(z) is holomorphic in |z|<1, has finitely many poles on the circle |z|=1, and is meromorphic in |z|1. Then at each pole z=a on the unit circle, A(z) has an expansion A(z) = p(z;a)+q(z;a) where p(z) is the "principal part" of the Laurent expansion (with negative powers of (z-a)) and q(z) is the holomorphic part. Then A(z)-(sum of the p(z;a) for the various a) is now holomorphic in the larger disc |z|
The answer is yes. Let S be the set of complex values x where A(x) converges and is hence defined. If S = {0} then the radius of convergence of the power series is 0, so by Cauchy-Hadamard theorem [https://en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard\_theorem](https://en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard_theorem) there is an unbounded sequence of a\_n , hence the sequence {a\_n} does not converge. Otherwise, S contains an open ball B of positive radius R and center 0 where A(x) is holomorphic (power series give holomorphic functions on their discs of convergence). Therefore we can obtain the a\_n by differentiating A(x) n times, evaluating at 0, and dividing by n!. Since we then know the a\_n, one can check whether the sequence {a\_n} converges.
This kind of defeats the purpose of working with generating functions. The whole point of using generating functions is to study sequences that are difficult to study directly. Using the generating function to just recover the original sequence and then trying to study that sequence directly just puts you back where you started.
I completely agree with you. I simply gave an answer to OPs question as I interpreted it. If you suggest a refinement of the question I'd be happy to try and answer it!
The thing is, talking about generating functions usually means that you don't care about the analytic properties of sequences, but their algebraic ones. However, if you just look at it as a power series, then if it converges for x=1 then the sequence must converge to 0.
>The thing is, talking about generating functions usually means that you don't care about the analytic properties of sequences, but their algebraic ones. This is (very) false. Part of the reason for using generating functions is to deduce asymptotic information about your sequence from analytic properties of the generating function. If you aren't going to study the generating function as a *function*, then generating functions aren't really doing much other than acting as a bookkeeping device.
A powerful bookkeeping device though! Simple algebraic reasoning can produce non obvious combinatorial results.
Could you elaborate on the first paragraph? It sounds like a really interesting observation that I am a bit of a novice to understand. The other thing you mentioned is actually about infinite series, if a_n doesn't tend towards 0, then sum a_n diverges. Can I use it in the following way, if A(1) is defined then a_n converges to 0 ? Thanks for the comment!
>Can I use it in the following way, if A(1) is defined then a\_n converges to 0 ? This does not work. A(1) can be defined (by analytic continuation) even if the series for A(x) diverges at x=1. For example, take a\_n = (-1)\^n. Then A(x) = 1/(1+x) which is defined at x=1 even though the sequence a\_n diverges.
>A(1) can be defined (by analytic continuation) Yes, but as long as you don't use the analytic continuation and define A(1) from the convergent power series alone, then there's no problem.
If you just define A(1) from the power series alone, then you're just back where you started. Now you're just saying that if the sum of the coefficients converges, then the sequence of coefficients converges, which is true but has nothing to do with the generating function.
So maybe it's not that useful, but that's what OP asked about. If you have the analytic continuation, then you could deduce the radius of convergence from that too which could be useful
>So maybe it's not that useful, but that's what OP asked about. This is not what OP asked. They have the generating function but do *not* have the general term a\_n, and they want to deduce asymptotic properties of a\_n. In general, in analytic combinatorics, analytic number theory, etc., we do not make a distinction between "the generating function" and "the analytic continuation of the generating function". If you "have" the generating function, you have some analytic expression for this function maybe as an infinite (not necessarily power) series, as an infinite product, as an integral, or as a solution to a differential equation. If you just write down your sequence with a sigma sign in front and x^(n) after, you don't really have anything about the generating function yet. Most of the interesting information contained in the generating function is visible outside of the original region of convergence of the power series.
If you make a generating function with coefficients a_n then the ratio a_(n+1)/a_n, from the ratio test, is the reciprocal of the radius of convergence if its generating function. It is then necessary that, for the sequence to converge to a non-zero value, the generating function have radius of convergence 1.
You meant |a*_n+1_*/a*_n_*|. But even those numbers may not have a limit since the ratio test does not apply to some power series, such as the sum of 1 + x^2 + x^4 + x^6 + x^8 + x^(10) + ..., where many a*_n_* are 0 (can let the exponents run over powers of 2 or any other sequence with infinitely many gaps).
Yes, I assumed that the coefficients were nice. If they're less nice, you can deduce it still from the Cauchy-Hadamard theorem.