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?
On
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 $k_n=2n+1$
On
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 $3$.
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.