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Assume we have the following series:

 1 - x^2/6 + x^4/120 - x^6/5040 + x^8/362880

Can I evaluate the radius of convergence for both of them? in Mathematica?

In general how can evaluate the radius of convergence is |an/an+1| for large n for any series in mathematica?

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  • $\begingroup$ I believe one way to estimate the radius of convergence is $|a_{n}/a_{n+1}|$ for large $n$. In this case, that limit goes to infinity, which would imply an infinite radius of convergence (which is true here, since I'm assuming that's the power series for $e^x$). $\endgroup$
    – march
    Commented Feb 26, 2020 at 16:41
  • $\begingroup$ Dear, yes I want to evaluate the radius of convergence is |an/an+1| for large n, for any series , what shall use in mathematica? $\endgroup$
    – user62716
    Commented Feb 26, 2020 at 16:45
  • $\begingroup$ In the new edit, you now have the first few terms of the Taylor series of cosine instead of the Taylor series for the exponential, but these two power series are intimately related to each other, and anyway, the first method I outlined below will work because we have a closed form for the coefficients of that entire series. I'm beginning to suspect a little that you need to read up a little on series, radius of convergence, Taylor series, and such. $\endgroup$
    – march
    Commented Feb 26, 2020 at 17:15
  • $\begingroup$ Thanks march do you have some good book or lecture notes for that!!! $\endgroup$
    – user62716
    Commented Feb 26, 2020 at 17:17

1 Answer 1

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One can either use SumConvergence or Sum with GenerateConditions -> True (thanks to a comment by Artes). As an example:

Sum[Cos[n] x^n, {n, 1, Infinity}, VerifyConvergence -> True, GenerateConditions -> True]
ConditionalExpression[-((x (1 + E^(2 I) - 2 E^I x))/(2 (E^I - x) (-1 + E^I x))), x != E^I && E^I x != 1 && Abs[x] < 1]

Unfortunately, SumConvergence doesn't seem to work:

SumConvergence[Cos[n] x^n, n]
SumConvergence[Cos[n] x^n, n, Assumptions -> -1 < x < 1]
SumConvergence[Cos[n] x^n, n, Method -> "RatioTest"]
SumConvergence[Cos[n] x^n, n, Method -> "RootTest"]
SumConvergence[Cos[n] x^n, n, Method -> "RaabeTest"]
SumConvergence[Cos[n] x^n, n, Method -> "IntegralTest"]

Below, I work through some of the theory and do this also numerically.


The radius of convergence $R$ is given by $$ \frac{1}{R} = \lim_{n\to\infty}\sup(\sqrt[n]{|a_n|}) $$ In this case, the coefficients are $\frac{1}{n!}$, which is a monotonically decreasing sequence, and so for our case, this becomes $$ \frac{1}{R} = \lim_{n\to\infty}\frac{1}{n!} $$ Using Mathematica,

Limit[1/Factorial[n], n -> ∞]
(* 0 *)

and hence the radius of convergence is infinity, as it should be for the Taylor series of the exponential function.


Here is an example with a more complicated function. Consider the series $$ \sum_{n=1}^{\infty}\cos(n)x^n. $$ In Mathematica, we get

Sum[Cos[n] x^n, {n, 1, \[Infinity]}] // ExpToTrig;
(Numerator@% Conjugate@Denominator@% // ComplexExpand)/( Denominator@% Conjugate@Denominator@% // ComplexExpand) // Simplify
(* (x (-x + Cos[1]))/(1 + x^2 - 2 x Cos[1]) *)

Plotting this function, we get

Plot[{(x (-x + Cos[1]))/(1 + x^2 - 2 x Cos[1]), Sum[Cos[n] x^n, {n, 1, 50}]}, {x, -1.01, 1.01}]

enter image description here

which suggests that the radius of convergence is 1. We can verify this numerically in the following way. Make a table of values of the coefficients:

tbl = Table[(N@Abs[Cos[n]])^(1/n), {n, 1, 10000}];

Then compute the maxima of successively larger subsets of the values and plot them:

Table[Max[tbl[[;; kk]]], {kk, 1, 10000}] // ListPlot

enter image description here

That sure appears to be equal to one!

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  • $\begingroup$ Thank you march, since it simple coefficients you can find it but for complicate series what shall we do for: 0.1 + 0.397953 t + 0.592849 t^2 + 0.588719 t^3 + 0.438295 t^4 + 0.260863 t^5 + 0.129195 t^6 $\endgroup$
    – user62716
    Commented Feb 26, 2020 at 16:51
  • $\begingroup$ @user62716. Perhaps give an example of a more complicated series and edit your OP with your true question! $\endgroup$
    – march
    Commented Feb 26, 2020 at 16:52
  • $\begingroup$ I modified the question, I want general way for any series, can we do it? $\endgroup$
    – user62716
    Commented Feb 26, 2020 at 16:55
  • $\begingroup$ But those aren't series, those are just polynomials. Unless you have a general expression for the (infinite number of) coefficients, then all you have are polynomials, and so the radius of convergence is of course always infinite. $\endgroup$
    – march
    Commented Feb 26, 2020 at 16:57
  • 2
    $\begingroup$ @user62716 What you are looking form is just SumConvergence, it has options where one can use e.g. "IntegralTest", "RaabeTest","RatioTest", however there is no tool which can predict another coefficient of a series you have in mind. $\endgroup$
    – Artes
    Commented Feb 26, 2020 at 17:00

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