# Convergence of a series that is only known numerically

I hope this is not a duplicate, this is quite a general question but I couldn't find the answer in any of the other posts.

Suppose one has a power series $$\sum_{n=0}^{\infty}c_n x^n$$ whose coefficients $$c_n$$ can only be numerically computed (for example, in my particular case, the $$c_n$$ are obtained by inverting a matrix). The general question here would be how to implement the usual convergence tests (ratio test and root test specifically) to extract whether the series has a finite or inifite convergence radius in $$x$$.

To be more specific on the problem of having access to only the numeric values of $$c_n$$, I'll show my attempts with the root and ratio tests. I compute the first 150 $$c_n$$ coefficients and plot $$\vert c_n\vert^{1/n}$$ as a function of $$n$$:

To me this picture does not make it clear whether or not the sequence approaches a finite value. With a little effort, I compute more coefficients, this time I go to 450. This is the result

Which looks as inconclusive as the previous picture, as I cannot tell if the sequence approaches a value at around $$2\pi$$ or grows to infinity.

Interestingly, the ratio test looks like this:

Where the sequence that I plotted is $$\vert c_{n+1}/c_n\vert$$. It seems that the "sub-sequence" given by odd n+1 and even n goes to zero while that given by even n+1 and odd n goes to infinity. I am unsure if the presence of the seemingly divergent contribution is a reason strong enough to claim the power series has zero radius of convergence.

In sum, I am having trouble extracting strong conclusions about the convergence radius of the power series from this kind of analysis, and I was wondering if there are methods that can be applied other than plotting a few terms and trying to guess the general behavior.

Edit: from the comments I see there is in principle no way of determining the convergence radius by looking at a finite number of terms. This is already an answer, but I'll leave here the specific problem that motivated the question.

Let us consider the matrix $$S$$, whose coefficients $$s_{n,k}$$ are

\begin{aligned} s_{n,k}&=0,\quad k

where

\begin{aligned} f(n,k)&=(2i\pi)^n \Gamma(k+1,-2 i \pi ,2 i \pi )\\& +(2i\pi)^{k+1}[(-1)^{k+n} (\Gamma (n,-2 i \pi )-\Gamma (n))+\Gamma (n,2 i \pi )-\Gamma(n)],\\ g(n)&=2\pi(\tilde{\Gamma}(n,-2 i \pi )+\tilde{\Gamma}(n,2 i \pi)-2)+i n \tilde{\Gamma}(n+1,2 i \pi,-2i\pi). \end{aligned}

In these expressions $$i$$ is the imaginary unit, $$\Gamma(z,a,b)$$ is the incomplete gamma function and $$\tilde{\Gamma}(z,a,b)=\Gamma(z,a,b)/\Gamma(z)$$.

The $$S$$ matrix is upper triangular, and all its diagonal elements are $$1$$. Let us now consider the inverse matrix $$T\equiv S^{-1}$$. Then the $$c_n$$ coefficients of the power series above are the elements of the first row of $$T$$.

In these conditions I have not been able to find a form for the entries $$t_{n,k}$$ of the $$T$$ matrix, and thus I've been forced to resort to generate $$S$$ numerically and invert it.

• all your plots are labeled "ratio test", which I think is a bit odd Commented Sep 30, 2022 at 10:02
• No amount of approximation will tell you information about a limit, unless you know additional information about what you are computing. If there is an original problem motivating this question, post that problem instead. Your modeling of it into this new question, makes it have the answer "nothing can be said".
– plop
Commented Sep 30, 2022 at 10:22
• Good question, but @user85667 has the correct answer here. Unfortunately without proper analytical bounds, we cannot say anything definitive about a sum when we can only access a finite number of its terms
– Zim
Commented Sep 30, 2022 at 10:32
• From the first plots it is quite logical to separate the odd and even index coefficients for the ratio test. Then the asymptotic behavior of the ratios $\frac{|c_{n+2}|}{|c_n|}$ should show a similar behavior to the root test (with a squared size, as 2 steps are combined). Commented Sep 30, 2022 at 11:49
• Yes, that remains the case. What you can do with such finite information is to build a model for asymptotic convergence, estimate its parameters, and then try to prove it via the sequence construction process. But for the overall question the answer appears quite clear, there is no convergence to zero visible in the graphs, so most likely the radius of convergence is finite. But obviously some slow divergence like in $\ln(\ln(n))$ and thus radius zero can not be excluded. Commented Sep 30, 2022 at 13:22

As it was pointed out in the comments, without further information, nothing can be said. What can you say about the coefficients $$c_n$$? Even without being able to compute them analytically, if you are able to show that $$|c_n| \leq a_n$$ and $$\sum a_n < \infty$$, you are done.