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I would like to compute the foillowing limit: $$\lim_{n \to \infty}\; \frac{1}{n \sin(\pi^6 n)}$$ assuming that $n$ is an integer number.

I use this code, but it does not give an answer.

Limit[1/(a Sin[a π^6]), a -> ∞, 
 Assumptions -> a > 0 && a ∈ Integers]
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    $\begingroup$ What you've done looks correct. Limit[1/(IntegerPart[a] Sin[IntegerPart[a] π^6]), a -> ∞] doesn't work either I'm afraid. $\endgroup$
    – flinty
    Commented Aug 31, 2020 at 22:06
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    $\begingroup$ You should see a warning: Limit::alimv "Warning: Assumptions that involve the limit variable are ignored." $\endgroup$
    – Bob Hanlon
    Commented Aug 31, 2020 at 22:23
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    $\begingroup$ By the way, this can almost certainly (in the colloquial rather than mathematical sense) be proved to not have a limit using Dirichlet's theorem on Diophantine approximation. $\endgroup$ Commented Sep 1, 2020 at 21:54
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    $\begingroup$ @user64494 No idea what you are going on about. Did you try to use the Dirichlet approximation result? On pi^5? You should be able to show that the value is infinitely often larger than pi or smaller than -pi, and infinitely often either the opposite or else close to zero. $\endgroup$ Commented Sep 2, 2020 at 16:11
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    $\begingroup$ @user64494 I would encourage you not to take cheap shots and make unclear comments. If you have to say or criticise something, give specific points and don't just say "no, you're wrong". If you have something valuable to add, then do it but refrain from making vague statements. The same is true for two of your comments under the answers given! $\endgroup$
    – halirutan
    Commented Sep 3, 2020 at 23:38

2 Answers 2

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An appropriate tool for calculating limits of sequences over integers is DiscreteLimit. However DiscreteLimit[1/(a Sin[π^6 a]), a -> ∞] cannot compute our task. On the other hand we can figure out that DiscreteLimit[a Sin[π^6 a], a -> ∞] yields Indeterminate, i.e. it says that the limit does not exist and. We can also find it calculating discrete limes superior and limes inferior:

DiscreteMaxLimit[ a Sin[π^6 a], a -> ∞]
DiscreteMinLimit[ a Sin[π^6 a], a -> ∞]
 ∞
-∞

Alternatively one can find it with standard limes superior and limes inferior, e.g. Through @ { MinLimit, MaxLimit}[1/(a Sin[π^6 a]), a -> ∞]. For an insight it is reasonable to plot appropriate sequence

DiscretePlot[ 1/(a Sin[π^6 a]), {a, 1000, 1240, 2}, ImageSize -> Large]

enter image description here

It is clearly seen that $\sin( \pi a)$ takes values between $-1$ and $1$, however $\sin(\pi^6 a)$ never equals but it can approach $0$ with a very good approximation for appropriately large integer values of $a$. E.g. we find $6$ values of $\sin(\pi^6 a)$ in the first $10^6$ natural numbers $a$ closest to $0$:

N[ TakeSmallestBy[ Sin[π^6 Range[10^6]], Abs, 6], 10]
 {-1.694781536*^-6, 3.389563072*^-6, -5.084344608*^-6, 6.779126144*^-6, 
  -8.47390768*^-6, 0.00001016868922}

and for our sequence they are

1/%
{-590046.5510, 295023.2755, -196682.1837, 147511.6377,
 -118009.3102, 98341.09183}
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 DiscreteLimit[(a Sin[Pi^6 a]), a -> Infinity]
                (*Indeterminate *)
    DiscretePlot[(a Sin[Pi^6 a]), {a, 1, 1000, 1}]

so $\lim_{n \to \infty}\; \frac{1}{n \sin(\pi^6 n)}$ divergence.

enter image description here

Or a method using series expansion

fx = Series[1/((1/x)* Sin[Pi^6 *(1/x)]), {x, 0, 1}] // Normal(* find the sereis expansion of it *)
Plot[fx, {x, -1, 1}]
Limit[fx, x -> 0]
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  • $\begingroup$ Hope you understand that the behavior of the first terms of a sequence says nothing about its limit at infinity. $\endgroup$
    – user64494
    Commented Sep 1, 2020 at 13:49
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    $\begingroup$ @user64494 But the plot gives a pretty good idea of what is happening. Proof might involve methods from Diophantine approximation (just guessing). $\endgroup$ Commented Sep 1, 2020 at 14:00
  • $\begingroup$ @Daniel Lightblau: This is rather a math question than a Mathematica question (see a related topic mathoverflow.net/questions/24579/convergence-of-sumn3-sin2n-1/…). Both the answers are primitive. $\endgroup$
    – user64494
    Commented Sep 1, 2020 at 15:36
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    $\begingroup$ @user64494 I would not call them primitive. Rather, they are uses of computation to explore the behavior. They give one incentive for attempting to prove the limit does not exist. (I realize experimental math is not to everyone's taste.) $\endgroup$ Commented Sep 1, 2020 at 21:48

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