Finding accumulation points of a set I am given the set $$E=\left\{\frac{n^{(-1)^{n}}}{n+1} \colon n\in\mathbb{N} \right\}$$ and asked to find all accumulation points (and prove so). I can prove that 0 and 1 are accumulation points and that for any real number $a$ where $$a<0 \qquad\text{or}\qquad a>1$$ that $a$ is not an accumulation point. However I'm having difficulty showing that there is no $$a \in (0,1)$$ that is an accumulation point. 
I might be able to show that no member of E itself is an accumulation point by induction but that seems awkward and clunky and it doesn't help me with all the other elements in (0,1). My instructor said this problem was supposed to be straightforward and I'm afraid that I've already overcomplicated it with so many cases on a. 
It seems intuitive that nothing in (0,1) is an accumulation point but I don't know how to construct an epsilon so that for any $$a\in (0,1)$$ that the epsilon neighborhood around this a necessarily has no element in A. 
I don't necessarily want a full solution. Some hint as to whether I actually need my cases, or how to go about constructing my epsilon is preferable.
 A: You can partition your set E into  $E_{2n}\cup E_{2n+1}$, and then you can see that 
$E_{2n}=\frac {n}{n+1}$ , which is strictly increasing, and $E_{2n+1}= \frac{1}{n(n+1)}$, which is strictly decreasing. This means that the limit points of E will be the LUB of $E_{2n}$ and the INF of $E_{2n+1}$ , which are $1$ and $0$ respectively, as you correctly pointed out (the only limit point of a bounded non-decreasing sequence of Reals is its LUB, a.k.a SUP, and the only limit point of a non-increasing sequence of Reals is its GLB a.k.a INF). This means that beyond a certain N, all your terms $E_{2n}$ will be within $e $ of 1, for any $e>0$, and all the odd terms $E_{2n+1}$ will be within $e'>0$ of $0$. This makes any possible third limit point $p$ impossible, as this potential third limit point cannot be approached within less than neither $e< \frac{1-p}{2}$ by points approaching 1, nor can it be approached closer than $e'<\frac{p}{2}$ by the terms approaching $0$, by choosing $N,N'$ large-enough for $E_{2n},E_{2n+1}$ respectively
A: Let $x_n=\frac {2n} {2n+1}$ and $y_n=\frac {\frac {1} {2n+1}} {2n+2}=\frac {1} {(2n+1)(2n+2)}$.  That is, $x_n$ and $y_n$ are $\frac {n^{(-1)^n}} {n+1}$ when we replace $n$ by $2n$ and $2n+1$, respectively.  From there, you can show that $x_n \to 1$ and $y_n \to 0$ and the result follows.
A: It's a different set but the process is the same and $\text{Acc}(\Omega)=\{-1,1\}\neq \text{Acc}(E)=\:\large?$ $$ \large\Omega\normalsize=\left\{\frac{{(-1)^{n}n}}{n+1}\:\colon n\in\mathbb{N} \right\}=\left\{{-1\over 2},{2\over 3},{-3\over 4},{4\over 5},...\right\}.$$
Let's conjecture that $${\large\Omega}^{C}=\underbrace{(-\infty,-1]}_{\mathbb R_{(-1)}}\bigcup_{n\in \mathbb N } \left [{-2n-1\over \:\:\:2n+2},{-2n+1\over \:\:\:2n}\right)\large \cup\normalsize (\frac{-1}{\:\:\:2},\frac{1}{2}\:)\bigcup_{n\in \mathbb N } \left ({2n-1\over 2n},{2n+1\over 2n+2}\right]\large \cup\normalsize\underbrace{[1,\infty)}_{\mathbb R_{(1)}}.\\ $$
Now, for the closed sets $\:\mathbb R_{(-1)}\&\:\mathbb R_{(1)},\:$we need only to look at the finite bounds to analyse  $\text{Acc}(\Omega)\:$because of the completeness property of $\:\mathbb R.$
In a rigourous fashion we can prove that because $\:\Omega\subset \mathbb Q,\:$ any neighbourhood $\:\mathcal N_{\delta_{>0}}\:$ around $-1\:\&\:1\:$ will be infinite in cardinality.
Therefore $\:(-1 \land 1)\in \text{Acc}(\Omega)\iff\{-1,1\}\subset \text{Acc}(\Omega)$.
Usually to prove that $\:\text{Acc}(\Omega)\subset\{-1,1\}\:$ we use absurdity. That is, we assume an arbitrary element $\:x\in\text{Acc}(\Omega)\:$ that must belong to $\Omega:$  $$\text{Let }\:\hat D=\Large|\normalsize{\frac{{|(-1)^{n+1}|(n+1)}}{n+2}}-\frac{{|(-1)^{n}|n}}{n+1}\Large|\normalsize\ > \frac{1}{(n+2)(n+1)}:=\hat\delta,\:\forall n\in\mathbb N \\ \text{Thus, }\:\exists\:\hat\delta_{_{>0}} <\hat D\:\:\:\text{s.t}\:\:\:\mathcal N_{\delta_{>0}}(x)\large\cap \normalsize \Omega=\emptyset \iff x\notin \text{Acc}(\Omega),$$
Contradiction.
Also note that $\:\overline \Omega\supset\text{Acc}(\Omega)=\{-1,1\}$
