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$$(4n-n^{2/5})/(4((1/n)-e^n))$$

I just need to know how to compute the subsequence in order to compute the cluster points.

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Can you be more specific? –  Spawn1701D Apr 11 '13 at 1:15
    
I was given the problem: compute the cluster points for the sequence (4n-n^(2/5))/(4((1/n)-e^n)). In order to compute a cluster point (or limit/ accumulation point) you take the limit of the subsequences. I need to understand how to find the subsequences or how to find the cluster points... –  kaitlyn Apr 11 '13 at 1:15
    
sorry! that was supposed to say "compute the subsequence" not computer! –  kaitlyn Apr 11 '13 at 1:20
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I think your question is more about mathematics than about Mathematica, which is a software program. If so, you should probably ask this on math.stackexchange.com –  Michael E2 Apr 11 '13 at 2:37
    
possible duplicate of Limit of a subsequence –  Artes Apr 11 '13 at 6:28
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closed as off topic by Michael E2, m_goldberg, Sjoerd C. de Vries, Artes, Yves Klett Apr 11 '13 at 6:49

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1 Answer

First define the sequence:

a[n_]= (4 n - n^(2/5))/(4 (1/n - Exp[n]))

its limit is

Limit[a[n],n->Infinity]

and gives 0. Now, if you want a specific subsequence of this sequence you can say

b[n_] = a[f[n]]

where $f$ can be for instance $f=2 n$ or $f=2n+1$.

Is this what you want?

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yes, I think that this is what I want. So my cluster/ limit point would be 0? how did you get f=2n and f=2n+1? –  kaitlyn Apr 11 '13 at 1:30
    
you just say a[2 n] or a[2n+1] –  Spawn1701D Apr 11 '13 at 1:34
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