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Determine all limits points of the following sequences:

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

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

  1. $n_k=2k$ 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.
  2. $n_k=2k+1$ 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:

  1. $n_k=2k$ 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.
  2. $n_k=2k+1$ 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:

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

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:

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

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

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It looks pretty nice and fine, yet I don't think you've explained in each case why there are no more limit points ... – DonAntonio Nov 25 '12 at 13:05
"We can definitely see, that |bn|≥0 and is therefore bounded": that's not right! – TonyK Nov 25 '12 at 13:07
@DonAntonio: Yeah thats one issue here I would like to solve, too. Any help is appreciated. – Christian Ivicevic Nov 25 '12 at 13:13
@TonyK: Could you give an example why this isn't true? – Christian Ivicevic Nov 25 '12 at 13:14
By the way, if you explicitly exhibit a limit point (as you do), there is no need to refer to Bolzano-Weierstrass to prove that one exists at all. – Hagen von Eitzen Nov 25 '12 at 13:42
up vote 1 down vote accepted

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. $_\square$

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You may try to prove the following easy proposition:

Proposition: Let $\,\Bbb N=A_1\cup\ldots\cup A_k\,$ be a partition of the naturals -- and thus $\,i\neq j\Longrightarrow A_i\cap A_j=\emptyset\,$ -- , and s.t. $\,\forall\,\,\,1\leq i\leq k\,\,\,,\,|A_i|=\aleph_0\,$ , and let $\,\{x_n\}_{n\in\Bbb N}\,$ be a real/complex sequence. Then

$$(1)\;\;\;\;\;\;\;\text{If there exist subsequences}\,\,\{x_{n_i}\}_{i\in A_i}\subset\{x_n\}_{n\in\Bbb N}\,\, \text{and a real number}\,\,L\,\,\,s.t.\,\,\,$$


$$(2)\;\;\;\;\;\;\;\;\text{If there exist subsequences}\,\,\{x_{n_i}\}_{i\in A_i}\subset\{x_n\}_{n\in\Bbb N}\,\,\,\, \text{and real numbers}\,\,L_i\,\,\,s.t.\,\,\,$$

$$\lim_{i\to\infty}x_{n_i}=L_i\,\,,\,\forall\,\,i=1,2,...,k\,\,\,,\,\,\text{then the set of all the limit points of}$$

$$\,\,\{x_n\}_{n\in\Bbb N}\,\,\,\text{is the set}\,\,\{L_i\}_{i=1}^k$$

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You seem to have found the limit points correctly, so I'll just talk about how to prove that that is all of the limit points possible. What you have shown in every question is that by breaking up the sequences into some sort of partition. i.e. even and odd terms, or terms equivalent $\mod 4$. The fact that you have partitioned the natural numbers is what you need to prove that you have found all the limit points.

For simplicity I'll talk about splitting into odd and even sub-sequences although an almost identical argument works for all parts.

Any sub-sequence must contain infinitely many odd terms of infinitely many even terms (or both). Now suppose we have a convergent sub-sequence. Then we can't have both infinitely many odd terms and infinitely many even terms, or we can find two sub-sequences which converge to different limits. (It is a well-known result that any convergent sub-sequence of a convergent sequence converges to the same limit.) Then we have finitely many even terms or finitely many odd terms. Hence after a certain point, the sequence becomes either all odd or all even, and so becomes a subsequence of one of the convergent sub-sequences you have already named. Hence it will converge to the same limit you have already named.

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