Trouble with the numerical evaluation of a series

Question

The series is $\;S=\displaystyle{\sum_{n=0}^\infty 2^{-n+\sin(n\pi/5)}}$.

Mathematica doesn't find a closed form for Sum[2^(-n + Sin[n Pi/5]), {n, 0, Infinity}], so I tried both:

N[Sum[2^(-n + Sin[n Pi/5]), {n, 0, Infinity}], 100]
NSum[2^(-n + Sin[n Pi/5]), {n, 0, Infinity}, WorkingPrecision -> 100]

They return the same result:

2.621953360503001622580428627210775515701951638633391919349668089988053230383633286891962065806603221

However, as I discovered later, this result is wrong. Indeed, the series is not difficult to evaluate, thanks to the periodicity of $\sin(n\pi/5)$:

$$S=\left(\sum_{n=0}^\infty2^{-10n}\right)\left(\sum_{n=0}^92^{-n+\sin(n\pi/5)}\right)=\frac{1}{1-2^{-10}}\sum_{n=0}^92^{-n+\sin(n\pi/5)}$$

And N[1/(1 - 2^-10) Sum[2^(-n + Sin[n Pi/5]), {n, 0, 9}], 100] yields:

2.621953365022156800044269458734842788555304465923212022625632939238430813027293562013936284925892294

So the initial sum only had 9 correct digits. It's so bad that I am almost certain I am missing an important option, but I have no idea which (I tried playing with AccuracyGoal and PrecisionGoal, but it led me nowhere so far).

Any idea to get the numerical answer directly?

The same method as above can be used to evaluate $\displaystyle{\sum_{n=0}^\infty 2^{-n+(-1)^n}}$, and the sum is $3$, however, Mathematica fails with a message that starts with NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections. So again I must be missing somthing important. The failure occurs with:

N[Sum[2^(-n + (-1)^n), {n, 0, Infinity}], 100]

My guess would be that it fails for this one because the correct answer is an integer, and thus it's impossible to give even a single correct digit since the sum could be 2.99999... or 3, but I'm not sure what really happens here, nor how to deal with this.

I'm not entirely new to Mathematica, but I have always found that numerical computations are difficult to handle with it. If someone can point to a good tutorial on this kind of problem or similar ones with NIntegrate, it would be really helpful!

Answer 1

What seems to help at least with NSum is increasing the number of terms that are computed explicitly via the option NSumTerms. This way one can get agreement for all first $100$ digits:

NSum[2^(-n + Sin[n Pi/5]), {n, 0, Infinity}, NSumTerms -> 400, WorkingPrecision -> 100]

2.621953365022156800044269458734842788555304465923212022625632939238430813027293562013936284925892294

N[1/(1 - 2^-10) Sum[2^(-n + Sin[n Pi/5]), {n, 0, 9}], 100]

2.621953365022156800044269458734842788555304465923212022625632939238430813027293562013936284925892294

Answer 2

Clear["Global`*"]

To find the number of terms required, find the minimum value of n for which the last term is less than 10^-100

min = MinValue[{n, 2^(-n + Sin[n Pi/5]) < 10^-100}, n, Integers]

(* 334 *)

sum1 = NSum[2^(-n + Sin[n Pi/5]), {n, 0, Infinity}, NSumTerms -> 400, 
  WorkingPrecision -> 100]

(* 2.6219533650221568000442694587348427885553044659232120226256329392384308130272\
93562013936284925892294 *)

sum2 = NSum[2^(-n + Sin[n Pi/5]), {n, 0, Infinity}, NSumTerms -> min, 
   WorkingPrecision -> 100] // Quiet

(* 2.6219533650221568000442694587348427885553044659232120226256329392384308130272\
93562013936284925892294 *)

sum1 === sum2

(* True *)

Answer 3

You can look at the Sin[] term in the exponent as some periodic noise. It would be good to smooth that out before trying numeric methods by summing over a period at a time:

sum = NSum[
  Sum[2^(-(10 n + k) + Sin[(10 n + k) Pi/5]), {k, 0, 9}],
  {n, 0, Infinity}, WorkingPrecision -> 100]
(*
2.62195336502215680004426945873484278855530446592321202262563293923843\
0813027293562013936284925892294
*)

1/(1 - 2^-10) Sum[2^(-n + Sin[n Pi/5]), {n, 0, 9}] == sum

(*  True  *)

Answer 4

If you compare for high m

m = 10^5; N[ParallelSum[2^(-n + Sin[n*(Pi/5)]), {n, 0, m*10 + 9}], 16]

with

N[Sum[2^(-10*n), {n, 0, m}]*Sum[2^(-n + Sin[n*(Pi/5)]), {n, 0, 9}], 16]

you get the same value. But if you then set m=Infinity the values differ, which is understandable, because any numeric sum up to infinity must include some approximation (or it would take forever...). You may notice that the calculation for m=10^5 takes much longer than the one up to Infinity.

Your second sum you can split into odd and even terms like to get

Sum[2^(-n + (-1)^n), {n, 0, Infinity}] = 
Sum[2^(-(2 n) + (-1)^(2 n)), {n, 0, Infinity}] + 
Sum[2^(-(2 n + 1) + (-1)^(2 n + 1)), {n, 0, Infinity}] = 3