Is there a way to evaluate the limit superior of a sequence? I didn't find any information about this in the documentation center.

The limit superior of a sequence $(x_n)$ is defined as

$$\limsup x_n=\lim_{n\to\infty}\sup\{x_k:k\ge n\}=\inf\{\sup\{x_k:k\ge n\}:n\in\Bbb N\}$$

  • 2
    $\begingroup$ Theoretically, Assuming[n ∈ Integers, Limit[MaxValue[{x[k], k >= n}, k, Integers], n -> Infinity]], but practically there are probably limitations depending on x. $\endgroup$ – Michael E2 Nov 1 '16 at 11:43
Max[Limit[expr, n -> Infinity]] 
(*expr is a function of n*)

I think you can try this because the limit superior of a sequence usually gives an interval. I don't have the privilege to comment here, therefore, I leave my suggestion for you.

  • $\begingroup$ The limit superior of a sequence is an unique value. I dont know why you says it is an "interval". More over: your solution dont work if the limit of the sequence doesnt exist. The limit superior of a sequence ever exists, but not the limit. Anyway I appreciate your help. $\endgroup$ – Masacroso Nov 2 '16 at 7:17
  • $\begingroup$ I think I should say that because the limit of a sequence from Mathematica (If it works.) probably gives an interval.(I guess) Then you just take the max value of the interval. ex. Limit[(-1)^n (1 + 1/n), n -> Infinity] gives (*E^(2 I Interval[{0, \[Pi]}])*) $\endgroup$ – tablecircle Nov 2 '16 at 8:36
  • $\begingroup$ Oh, I understand... thank you very much. I will try to know this behavior of mathematica. $\endgroup$ – Masacroso Nov 2 '16 at 8:45

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