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i am tryint to find a cluster point of this sequence, but i am having difficulties in definitions.

the sequence is this: $(a_{n})_{n \in \Bbb{N}}$ with $a_{n}:=(2+(-1)^n)\frac{n}{n+1}$

the definition of a cluster point is: there is some number as a cluster point to which the subsequence of a sequence converges to.

but how to find them? there are many right?

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As $n$ increases, $\frac{n}{n+1}$ approaches $1$, so $a_n$ is close to $2+1=3$ for large even $n$ and close to $2-1=1$ for large odd $n$. Thus, the cluster points must be $1$ and $3$, and all that remains is to prove that rigorously. This entails two separate tasks: you must show that $1$ and $3$ are cluster points of the sequence, and you must show that nothing else is a cluster point of it.

For the first task, just find subsequences converging to $1$ and to $3$; if you think about how I found $1$ and $3$ in the first place, this should be pretty easy.

For the second, let $x$ be any real number different from $1$ and $3$, and find an open interval around $x$ that contains only finitely many terms of the sequence. This is easy if you make sure that the interval does not have $1$ or $3$ in its closure.

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    Thanks Brian. brilliant explanation. So the cluster point is just the limit of a sequence , right?2012-12-22
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    @doniyor: Not quite. If a sequence actually converges to a limit, then that limit is its only cluster point. In general, though, the cluster points are the points that are limits of subsequences, so you can have lots of them.2012-12-22
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    So, the cluster point is small limits of subsequences of main sequence. the main sequence has only ONE cluster point and subsequences can have more than one, since there can be many subsequences.. right?2012-12-22
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    @doniyor: No, if the sequence actually converges, then **every** subsequence will converge to the same limit. It’s only when the sequence **doesn’t** converge that you can get subsequences that have different limits, as in the example in your question.2012-12-22
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    here is the clue that i didnot get before, thanks a lot!2012-12-22
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    @doniyor: You’re welcome!2012-12-22
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Consider the subsequence $k_n=2n$ (of even natural numbers). Then $$a_{k_n}=(2+1)\frac{2n}{2n+1}=6\frac{1}{2+\frac1n}\to\frac{6}{2}=3$$ How do we choose the subsequence? Well, the bad term here is $(-1)^n$ and we want to remove it from our expression. We could have also considered $k_n=2n+1$

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    Thanks Nameless. to "how do we choose the subsequence?": we just take any sequence out of the main sequence that the taken subsequence should converge to $3$ right?2012-12-22
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    @doniyor Pretty much. You have a freedom of choosing the subsequence you want2012-12-22
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    okay cool, thanks2012-12-22
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Sometimes expanding a sequence may be helpful to visualize its cluster points. Such as if we write the given sequence as:

{$[2-1]{1\over 2},[2+1]{2\over 3},[2-1]{3\over 4},[2+1]{5\over 4},...$} i.e. as {${1\over 2},3.{2\over 3},{3\over 4},3.{4\over 5},...$}

So {${1\over 2},{3\over 4},{5\over 6}...$} & {$3\times {2\over 3},3\times {4\over 5},3\times {5\over 6},...$} are the only [why?] convergent subsequences of {$a_n$} where the first one converges to $1$ & second one to $3$.