# Wolfram Language 12 says this absolutely convergent series does not converge. Is there any similar example?

I am reading "Lectures on complex function theory" by Takaaki Nomura.
In this book, there is the following example:

$$\sum_{n=1}^{\infty} \sin(\pi(2+\sqrt{3})^n)$$ converges absolutely.

But Wolfram Language 12 says this absolutely convergent series does not converge. Is there any similar example?

Is there any reason why Wolfram Language 12 says this series does not converge?

• math.stackexchange.com/questions/835554/… Feb 3 '20 at 14:27
• math.stackexchange.com/questions/97548/… Feb 3 '20 at 14:30
• $\sum \sin(\pi\,\alpha^n)$ for a quadratic integer $\alpha = a+\sqrt{b}$ will be absolutely convergent whenever $|\bar\alpha| = |a-\sqrt{b}| < 1$ because $\alpha^n+\bar\alpha^n$ is a rational integer. (It may be convergent/divergent for other $\alpha$, but this applies to $\alpha = 2 \sqrt{3}$, the golden ratio, etc.) Feb 4 '20 at 12:43
• $\sum \sin(\pi\,\alpha^n)$ often diverges, doesn't it? It makes the OP's example seem a special case that M does not handle. It also fails on similar algebraic integers with conjugates less than 1 in absolute value, such as $\alpha = {\sqrt[3]{9-\sqrt{69}}+\sqrt[3]{9+\sqrt{69}}}/{\sqrt[3]{18}}$. Feb 4 '20 at 13:31
• Oops, typos: $\alpha = 2 + \sqrt{3}$, ${\left(\sqrt[3]{9-\sqrt{69}}+\sqrt[3]{9+\sqrt{69}}\right)}\big/{\sqrt[3]{18}}$. Feb 4 '20 at 14:40

The references from wuyudi seem meaningful but not very targeted. The attempt by Cesareo is probably a good ansatz, but not a solution. The sum of 2 + Sqrt1 is bigger than 1 and the series in the argument fairly divergent.

A proper plot will give a proximate solution in the limits of the numerical accuracy of the local Mathematica installation.

You are in need of numerical experimentation first. Use NSum and You will get a numerical instability message for

ListPlot[Table[
NSum[Sin[\[Pi] (2 + Sqrt[3])^i], {i, 1, k}], {k, 1, 45}]]


Euler-McLaurin sum failed to converge to requested error tolerance.

The hypothesis of absolute convergence might hold despite that.

The series converges rapidly to the value of -1.05201 for my local presicion for the input

NSum[Sin[\[Pi] (2 + Sqrt[3])^i], {i, 1, \[Infinity]}]


The page of reference is Numerical Evaluation of Sums and Products.

Mathematica uses the method "EulerMaclaurin" even if I did request it under the more general preset option Automatic.

Despite some regions for n is which the series of the sums value behave very constant there are regions of wild oscillations. This is dependent on how big the sin is and how the last summation value is negative of positive.

The table

Table[NSum[Sin[\[Pi] (2 + Sqrt[3])^i], {i, 1, Infinity},
Method -> "EulerMaclaurin", VerifyConvergence -> True,
NSumTerms -> n, WorkingPrecision -> 10], {n, 25, 250, 25}]


Delivers

{0.*10^-10, 0.*10^-10, 0.*10^-10, 0.*10^-10, 0.*10^-10, 0.*10^-10,
0.*10^-10, 0.*10^-10, 0.*10^-10, 0.*10^-10}


That seems to be the correct result for the infinite series, but the is a message:

The series seems to be related to the SinIntegral built-in function of Mathematica. The relation is done via the Euler-Maclarin represenation of the Sin and then strait forward. It seems computational hard for Mathematica to do this symbolical.

More help may be possible on other pathes.

To get some understanding of what happens and stick close to the Mathematica corpus a probable idea is this entree question, making this question somehow kind of duplicate,

Paths integrals in the complex plane

This question has a different starting point so this connection between both question is hard for newcomers. Since Mathematica is not a kind of complex function theory specialist software package even more complex question may arise.

This given sum is one with the trigonometric function Sin. It can be represented with Exp[i x] with i the imaginary unit. Like Mr. Lichtblau mentioned in his answer.

The path in the integral is this Exp[i x] and has to parametrized correspondingly.

Next step is to concern oneself that with the exponential function this does not get very simplistic. It is the path parametrization for the residue that can be used to calculate the result in a symbolic fashion.

So it is hard to find a nice symbolic representation of the residue in the Mathematica help and on mathematica.stackexchange.com this is somehow ground breaking stuff.

We are now operating with the definition from:

We make use of the part "Calculating residue" from this definition. Most newcomers are already past away at this pace. It needs months of introductory universitary lessons to cope and hold the pace.

We like to discard the terms in the infinite sum that do not contribute to the sum at all. So only the terms without a pole contribute. That are not many. A nice and short help can be

Complex Residue by Eric Weisstein,

but must not be.

All that is used at last is, this is a holomorphic function in all terms and integrated in the residue to zero caused be that.

Since we used the Euler-Maclaurin formula to represent the Sin we have to take the imaginary part of our result. That is zero too and everything is proved.

The main part of the prove is the criteria absolute convergence of the series summed over and the sums too. It does not matter wether the resulting value is positive or negative in finite sum approximations. The only argument is finite.

Mathematica regulate its value internally so I did not manage to make it either only approximation or symbolic calculations. At last the NSum calculations did choose the infinite sums despite preselecting a numerical approximation. So all values in the table given are correct zero.

Like the definitions Contour integration show the solution is very easy understandable for newcomers.

The Pi(2+Sqrt3)^i indicate just how many time the pole or no pole is curved around by the integrating paths in the complex plain.

Take care not to switch wrong Exp[x^i] is unequal Exp[i x] and even Exp[x]^i for simplicity.

Pi(2+Sqrt3)^i is getting greater with i and so the path is getting longer only taken into account by the path integration in the path integral interpretation of the integral, see Answer of Mr. Weisstein already referenced.

Hope that helps and shows up, this is much to be written and an example better left in the introductory book. Mathematica can deal with it in version 12, but it is hard work making it do the trick with a stepwise solution. I do not know if there is a Wolfram Alpha specialization with more help. The one I tried had no trial and was not even accessable.