TITLE: Determining limit points and proving there are no more of them
QUESTION [6 upvotes]: Determine all limits points of the following sequences:

$\displaystyle a_n=\begin{cases}2^{-n},&n\text{ even},\\3^{1/n},&n\text{ odd}.\end{cases}$
$\displaystyle b_n=n+\frac{2(-1)^nn^2+3}{2n+1}$
$\displaystyle c_n=\sin\left(\frac{n\pi}{4}\right)$
$\displaystyle d_n=\frac{n+1}{n}\cdot i^n$

Explain for each sequence why the specified points are limit points and why there are no other limit points except than the ones you determined.

I am famous for doing everything way too difficult, so I would both like to know whether my results are correct and how to improve them considering length and comprehensibility. Furthermore I don't know how to prove that I found all limit points - any help/hints?
For the proof that there is at least one limit point I use the Bolzano-Weierstrass theorem which states that every bounded sequence has at least one converging subsequence.

Sequence 1
Just by looking at the sequence we see, that $|a_n|\le\sqrt{3}$ and therefore at least one converging subsequence with the related limit points does exist. We take subsequences $(a_{n_k})_{k\in\mathbb{N}_0}$ with

$n_k=2k$ such that $\displaystyle(a_{2k})_{k\in\mathbb{N}}=\frac{1}{2^{2k}}$ and we know that this is a null sequence and $0$ is therefore one limit point.
$n_k=2k+1$ such that $\displaystyle(a_{2k+1})_{k\in\mathbb{N}}=\sqrt[2k+1]{3}$ and we know that the $n$-th square root is convergent and therefore $\lim_{n\to\infty}\sqrt[n]{3}=1$ which is the second (and last) limit point.

Sequence 2
We can definitely see, that $|b_n|\ge 0$ and is therefore bounded and there exsists at least one limit point. We do look once again at subsequences $(b_{n_k})_{k\in\mathbb{N}_0}$ with the following two cases:

$n_k=2k$ such that we get $$2k+\frac{2(2k)^2+3}{2(2k)+1}=2k+\frac{8k^2+3}{4k+1}=\frac{16k^2+2k+3}{4k+1}\longrightarrow +\infty$$ which is no limit point.
$n_k=2k+1$ such that $$2k+1+\frac{-2(2k+1)^2+3}{2(2k+1)+1}=2k+1+\frac{-8k^2-8k+1}{4k+3}=\frac{k(2+4/k)}{k(4+3/k)}\longrightarrow \frac{1}{2}$$ which is our one and only limit point for this sequence.

Sequence 3
The basic properties of $\sin$ imply that $\displaystyle\left|\sin\left(\frac{n\pi}{4}\right)\right|\le 1$ and we should find some limit points. In fact we will find five of them. Assuming that we know that $\sin(0)=0$, $\cos(0)=1$, $\sin(\pi)=1$, $\cos(\pi)=-1$, $\sin(\pi/4)=1/\sqrt{2}$, $\sin(-\pi/4)=-1/\sqrt{2}$, $\sin(x+2k\pi)=\sin(x)$, $\cos(x+2k\pi)=\cos(x)$ and $\sin(x+\pi/2)=\cos(x)$ we again do look at subsequences $(c_{n_k})_{k\in\mathbb{N}_0}$ where we distinguisch between 8 cases:

$n_k=8k:$ $$\sin\left(\frac{8k\pi}{4}\right)=\sin(2k\pi)=0$$
$n_k=8k+1:$ $$\sin\left(\frac{(8k+1)\pi}{4}\right)=\sin\left(\frac{\pi}{4}+2k\pi\right)=\sin\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$$
$n_k=8k+2:$ $$\sin\left(\frac{(8k+2)\pi}{4}\right)=\sin\left(\frac{(4k+1)\pi}{2}\right)=\cos(2k\pi)=\cos(0)=1$$
$n_k=8k+3:$ $$\sin\left(\frac{(8k+3)\pi}{4}\right)=\sin\left(\frac{3\pi}{4}+2k\pi\right)=\sin\left(\frac{3\pi}{4}\right)=\sin\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$$
$n_k=8k+4:$ $$\sin\left(\frac{(8k+4)\pi}{4}\right)=\sin\left((2k+1)\pi\right)=\sin(\pi)=0$$
$n_k=8k+5:$ $$\sin\left(\frac{(8k+5)\pi}{4}\right)=\sin\left(\frac{5\pi}{4}+2k\pi\right)=\sin\left(\frac{5\pi}{4}\right)=-\frac{1}{\sqrt{2}}$$
$n_k=8k+6:$ $$\sin\left(\frac{(8k+6)\pi}{4}\right)=\sin\left(\frac{(4k+3)\pi}{2}\right)=\sin\left(2k\pi+\frac{3\pi}{2}\right)=\sin\left(\frac{3\pi}{2}\right)=\cos(\pi)=-1$$
$n_k=8k+7:$ $$\sin\left(\frac{(8k+7)\pi}{4}\right)=\sin\left(-\frac{\pi}{4}+2k\pi\right)=\sin\left(-\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}$$

Therefore we have five limit points: $0,\pm 1,\pm 1/\sqrt{2}$.
Sequence 4
This sequence is apparently bounded by the unit circle and therefore we know that $|d_n|\ge 1$. We distinguish here four different cases:

$\frac{4k+1}{4k}\cdot 1\longrightarrow 1$
$\frac{4k+2}{4k+1}\cdot i\longrightarrow i$
$\frac{4k+3}{4k+2}\cdot (-1)\longrightarrow (-1)$
$\frac{4k+4}{4k+3}\cdot (-i)\longrightarrow (-i)$

This leads to the result that each $i^k$ is a limit point.

REPLY [4 votes]: Let a sequence $(x_n)_{n\in N}$ be given.
Definition. If  $A\subseteq \mathbb N$ an infinite subset, let $L(A)$ be the set of all limit points of $(x_n)_{n\in A}$.
Proposition. If $\mathbb N=A\cup B$ where $A,B$ are infinite, then $L(\mathbb N)=L(A)\cup L(B)$.
Proof: Assume $a\in L(\mathbb N)$. Then for each $\epsilon>0$ there are infinitely indices $n$ with $|x_n-a|<\epsilon$. By the pigeon-hole principle, at least one of $A,B$ must contain infinitely many of these indices, i.e. $a$ is also a limit point of the corresponding subsequence, hence $a\in L(A)$ or $a\in L(B)$.
On the other hand, $a\in L(A)$ or $a\in L(B)$ clearly implies $a\in L(\mathbb N)$. $_\square$
Corollary. Assume $\mathbb N=A_1\cup \cdots \cup A_m$ where the $A_i$ are infnite and for $1\le i\le m$, the subsequence $(x_n)_{n\in A_i}$ converges to some limit $a_i$ (or diverges definitely). Then $L(\mathbb N)=\{a_i\mid 1\le i \le m\}$.
Proof: The step from two to $m$ sets is just induction. For the rest, note that the limit of a converging sequence is its only limit point. $_\square$