The Taylor expansion of function is very useful, but the convergence radius of power series results after Taylor expansion is also important. But the result of Series function does not contain the information of convergence region. What should I do to get the information of convergence region at the same time?

Examples for testing:

    Series[Power[Log[1-x]^2, (2)^-1]/Sqrt[x],{x,0,5}]
    Series[Sin[m*x], {x, 0, 5}]
  • $\begingroup$ For a function that is complex-differentiable in a neighborhood of the center of the series, the radius is the distance from the center to the nearest singularity. Your first example however is not complex-differentiable at zero, and as a function of a real variable, it is not defined in a neighborhood of 0. $\endgroup$
    – Michael E2
    Commented Aug 12, 2020 at 14:26
  • $\begingroup$ @MariuszIwaniuk Thank you for your help. Although the post you provided is similar to my question, this post has no adopted answer. $\endgroup$ Commented Aug 12, 2020 at 22:16

1 Answer 1


In principle, the formula (for example, see the Wiki article) $$ \frac 1 R=\limsup_{n->\infty} |a_n|^{\frac 1 n}$$ together with the SeriesCoefficint command should do the job for power series. However, there are technical problems with it.

(i) Series[Power[Log[1-x]^2, (2)^-1]/Sqrt[x],{x,0,5}] performs

$$ O\left(x^{11/2}\right)+\frac{\sqrt{x} \sqrt{x^2}}{x}+\frac{\sqrt{x^2} x^{3/2}}{2 x}+\frac{\sqrt{x^2} x^{5/2}}{3 x}+\frac{\sqrt{x^2} x^{7/2}}{4 x}+\frac{\sqrt{x^2} x^{9/2}}{5 x}.$$ This is not a power series. Moreover, the branch cut along the negative ray should be taken into account.

(ii)The complicated formula for SeriesCoefficient[Tan[x], {x, 0, n}] $$\begin{cases} \frac{i^{n+1} 2^n \left((-1)^n-1\right) \left(2^{n+1}-1\right) B_{n+1}}{(n+1)!} & n\geq 1 \\ 0 & \text{True} \end{cases}$$ (see the Wiki article) is implemented in Mathematica. However, it is difficult to operate with that:

a = Assuming[n >= 1,ComplexExpand[Abs[SeriesCoefficient[Tan[x], {x, 0, n}]]]]
(*(1/Sqrt[(((1 + n)!)^2)])2^n Sqrt[(-1 + 2^(1 + n))^2] Sqrt[Im[BernoulliB[1 + n]]^2 + 
Re[BernoulliB[1 + n]]^2] Sqrt[(1 + Cos[(1 + n) \[Pi]])^2 +  Sin[(1 + n) \[Pi]]^2]*)
b = FullSimplify[a, Assumptions -> n >= 1]
(*(2^(1/2 + n) (-1 + 2^(1 + n)) Abs[BernoulliB[1 + n]] Sqrt[ 1 + Cos[(1 + n)\[Pi]]])/Gamma[2 + n]*)


DiscreteMaxLimit[b^(1/n), n -> Infinity]

is running without any result during a long time.

(iii) Here I succeed for concrete values of $m$, e.g.

a[n_] := SeriesCoefficient[Sin[(3 + I)*x], {x, 0, n}, Assumptions -> n > 0]
Assuming[n >= 1, ComplexExpand[Abs[a[n]]]], 
Assumptions -> n >= 1])^(1/n), n -> Infinity]

This means the power series converges for each complex number $x$. It should be noticed that

SumConvergence[SeriesCoefficient[Sin[3*x], {x, 0, n}, Assumptions -> n > 0]*x^n, n]

returns the input.


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