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.