Prove that if $\left\{ x_n \right\}$ is an infinite sequence of real numbers, $x \in \mathbb{R}$, and every subsequence $\left\{ x_{n_k} \right\}$ has a subsequence $\left\{ x_{n_{k_j}} \right\}$ with $x_{n_{k_j}} \rightarrow x$, then $x_n \rightarrow x$.   I know that if every subsequence of a sequence converges to the same number, then the sequence converges to that same number. But I don't know if the same can be applied to subsubsequences. So for this problem, can I safely state that because $x_{n_{k_j}} \rightarrow x$, it is also true that $x_{n_k} \rightarrow x$? If this is true, then does that mean every subsequence $\left\{ x_{n_k} \right\}$ also converges to $x$?
23431
Is the following true? Let $x_n$ be a sequence with the following property: Every subsequence of $x_n$ has a further subsequence which converges to $x$. Then the sequence $x_n$ converges to $x$.  I guess that it is true but I am not sure how to prove this.
46326
I have a positive infinite sequence $a_n$ and I know that $a_n$$a_{n+1}$ converges to 1 and also $a_n$$a_{n+2}$ converges to 1.  Why does it mean that $a_n$ also converges to 1? How can I prove it?  Thanks.
99650
$(\to)$ Assume that $S \subseteq R$ is bounded. If $(x_n)$ is a sequence in $S$ , then $(x_n)$ is bounded and thus it has a convergent subsequence.  $(\leftarrow)$ Assume that $S$ is not bounded. Then for any integer $n$ there is an $x_n$ in $S$ such that $|x_n| &gt; n$. Since any convergent sequence is bounded, the sequence $(x_n)$ cannot have a convergent subsequence.  Can someone check to see whether my proof is missing something.  thank you.
122061
Let $a_n$ be an infinite sequence. The limit of the difference of two consecutive members is equal to $0$. Can we conclude that the sequence itself has a limit?  My attempt:  We have  $$ \lim_{n\rightarrow\infty}{a_n} - \lim_{n\rightarrow\infty}{a_{n-1}} = 0 $$ since as $n$ approaches infinity $a_{n-1}$ gets arbitrarily close to $a_n$ the sequence cannot diverge or be bounded but have no limit.  Is my proof correct and how would I be able to formalize the last sentence? Thanks
27721
Let the sequence be defined recursively $X_{n+1}=pX_{n}+q$ for p nonzero and $X_{1}$ arbitrary. In what conditions does the sequence converge?  I bumped into this problem and I don't know where to even begin. I would really appreciate some help.
90392
If $x_1$, $x_2$ are arbitary real numbers, $\alpha\in (0,1]$ and  $x_n=\alpha x_{n-1}+(1-\alpha)x_{n-2}$ for every positive integer $n$ (>2), show that the sequence $\{x_n\}$ is convergent. (Given $x_1&lt;x_2$)  I cannot rearrange to test convergency. Please help me by solving the problem.   Edit  I want to prove by using the property below:     If $\{x_{2n}\}$ and $\{x_{2n-1}\}$ converges to same limit $l$ then $\{x_{n}\}$ converges to $l$.
10491
I have a question regarding one exercise in Stephen Abbotts' Understanding Analysis. The question is: Assume $(a_n)$ is a bounded sequence with the property that every convergent subsequence of $(a_n)$ converges to the same limit $a \in \mathbb{R}$. Show that $(a_n)$ must converge to $a$.  I was then presented this proof (from the solutions manual): Assume for contradiction that $(a_n)$ does not converge to $a$, then from the negation of convergence we have (I'm not quite sure in this part): $$\exists \epsilon&gt;0, \forall N\in \mathbb{N}, \exists n\in \mathbb{N} : n\geq N \wedge |a_n-a|\geq \epsilon. $$  Using this we can construct a subsequence $(a_{n_j})$ that diverges from $a$ as follows: for arbitrary $N\in \mathbb{N}$, we can always find a, say, $n_1$ such that, $n_1 \geq N$ where $|a_{n_1}-a|\geq \epsilon$. Since $n_1$ itself is in $\mathbb{N}$, then we can again find a $n_2 \geq n_1$ so that $|a_{n_2}-a|\geq \epsilon$. And in general we can find a $a_{n_{j+1}}$ after choosing an appropriate $a_{n_j}$ so that $|a_{n_{j+1}}-a|\geq \epsilon$.  From the construction of such a subsequence, isn't the contradiction that $(a_{n_j})$ does not converge to $a$ already proof that $(a_n)$ should converge to $a$? I am asking because the proof that I've read continues to say that the Bolzanno-Weierstrass theorem can be used to get another subsequence from the constructed $(a_{n_j})$ which diverges from $a$, and that the proof ends there. I think I understand it, but I fail to see why it is necessary. Can somebody enlighten me on this?
68003
When selecting the terms of subsequence from each bisections, I thought axiom of choice might be required. But I'm not so sure whether or not, so please tell me. [edited] I'm sorry for the lack of explanation. I want to prove this statement: Let $a_1, a_2, \ldots \in \mathbf{R}$, and $(a_n)_{n\in\mathbf{N}}$ is bounded, then $(a_n)$ has some convergent subsequence.  The proof is as follows. Since $(a_n)$ is bounded, for all $n\in\mathbf{N}$,   $a_n \in I = [b, c]$.  Now, let $I_0 = I$ and if $I_n = [b_n, c_n]$,  we define $d_n = (b_n+c_n)/2$ and  if infinite terms of $(a_n)$ is included in $[b_n, d_n]$(resp. $[d_n, c_n]$), we will define  $I_{n+1} = [b_n, d_n]$(resp. $[d_n, c_n]$).If both intervals contain infinite terms, let $I_{n+1}$ be $[d_n, c_n]$. For all $n\in \mathbf{N}$, infinite numbers of $m \in \mathbf{N}$ exist such that $a_m \in I_n$ suffices. We take the sequence of natural numbers $(n(k))_{k\in\mathbf{N}}$which suffices $n(0) &lt; n(1) &lt; \cdots &lt; n(k) &lt; \cdots$ following this procedure: Now we have already selected $a_{n(1)}, \ldots, a_{n(k)}$, there are infinite numbers of $m\in \mathbf{N}$ which suffices $n(k)&lt;m, a_m \in I_{k+1}$, so let's take the minimum m out of it. Applying this process recursively, we obtain a infinite convergent subsequence(?). I think intuitively, by only repeating this process we can't obtain countable infinite terms of subsequence because we have to repeat infinite times.
2069
Question: Consider a sequence $\{a_{n}\}$ such that $a_{1},a_{2}&gt;0$, and for all $n \in \mathbb{N}$ we have:  $$a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}$$  Prove that :  $\displaystyle\lim_{n\to\infty}a_{n}$ exists and find this limit.  My work: If this limit exists, let $\displaystyle\lim_{n\to\infty}a_{n}=x&gt;0$; then we have $$x=\sqrt{x}+\sqrt{x}\Longrightarrow x=4$$ But I can't see how to prove the limit of $\{a_{n}\}$ exists. Thank you for you help!
79361
Show that $a_n=\sin(n)$ does not converge     My idea:  Take two subsequences: $a_{n_k}=\sin(\frac {\pi k} 2)$ , $a_{n_l}=\sin(\frac {2\pi l} 3)$  So: $\forall n$ : $\lim_{n\to\infty} a_{n_k}=1$, $\lim_{n\to\infty} a_{n_l}=-1$  So two inifinite subsequence converge to different limits thus the sequence $a_n$ doesn't converge.   Is that correct ?  Edit:   The contrapositive of the defintion of a limit of a sequence:   $\exists\epsilon&gt;0 : \forall n\in N : \exists n&gt;N \Rightarrow |x_n-L|&gt;\epsilon$  Take $\epsilon =1$ and we know that $\sin(n)$ is bounded so lets take it's supermum: 1  $|1-1|&gt;\epsilon=1\Rightarrow 0&gt;1 \Rightarrow$ Contradiction.
47059
I have tried to prove the following result, but it looks a bit messy in my opinion. How can it be improved?     Proposition. Let $a_n$ be a bounded sequence with values in $\mathbf{R}$.   Let $A=\{l\in\mathbf{R} \mid \text{ there exists a subsequence of $a_n$ &gt;converging to }l\}$. Then $A$ is closed.   Lemma: For each subsequence of $a_n$ which converges to $l$ it is true that $\forall {\varepsilon}&gt;0$ $a_n\in B(l,{\varepsilon})$ for arbitrarily large $n$ -- that is, for each $m\in\mathbf{N} \; \exists M\geq m \mid a_M\in B(l,{\varepsilon})$.  Let $a_{k_n}$\,be such a subsequence. By definition, for each $\varepsilon&gt;0 \; \exists N$ such that $\forall n\ge N \; a_{k_n}\in B(l,\varepsilon)$. Let $m\in\mathbf{N}$. Then $k_m\ge m$, since $a_{k_n}$ is a subsequence of $a_n$.   if $m\ge N$, take $M:=k_m$; if $m&lt;N$, take $M:=k_N$.   Proof. Let $x$ be a closure point of $A$. By definition, $\forall \varepsilon&gt;0 \; B(x,\varepsilon)\cap A\neq\varnothing$, which means that $\forall \varepsilon&gt;0$ there exists a subsequence $a_{k_n}$ which converges to ${l}_\varepsilon\in B(x,\varepsilon)$.  This means in particular that for all $m\in\mathbf{N}$ there exists a subsequence $a_{k^{(m)}_n}$ that converges to $l_m\in B(x,1/2^{m+1})$ .  Define the sequence $x_n$ inductively:   $x_0=0$; $x_n=\min\{m&gt;x_{n-1} \mid a_m \in B(l_m,1/2^{m+1}) \}$.    The sequence is well-defined, because there exists a subsequence converging to $l_m$, and by the lemma there are infinite indices for which the sequence lies in the neighborhood. So the set is not empty and by well-ordering there exists a minimum element.  It follows that for each $n\in\mathbf{N}$ $$x-1/2^{m}&lt;l_m-1/2^{m+1}&lt;a_{x_n}&lt;l_m+1/2^{m+1}&lt;x+1/2^{m},$$  so by squeeze theorem (strict inequalities imply non-strict inequalities) $a_{x_n}\to x$.
110106
x is a real number between 0 and 1  $ \lim_{n\rightarrow\infty} x ^ {\left(\frac 1 n\right)} = 1 $  SO far i know you need to do this to prove it:  let b>1, prove that $\root n \of b \rightarrow 1$ as $n \rightarrow\infty$  I was given this hint: let $a_n = \root n \of b - 1$ , prove that  $a_n \rightarrow 0$  Thanks!
14786
I'm not entirely sure how to go about proving this so hopefully someone can point me in the right direction. The definition I have for a limit point is "$a$ will be a limit point if for a sequence $x_n$ there exists a subsequence $(x_{n_k})$ such that $\lim_{k\to\infty} (x_{n_k})=a$".   In the forward direction if $x_n$ converges to a value $a$ then every convergent subsequence of $x_n$ must also converge to $a$. I'm not entirely sure where else to go from here. Intuitively I know that what I'm trying to prove must be true since if $x_n$ converges to $x$ then there exists an $N\in\mathbb{N}$ such that $\forall n\geq N$ we have infinitely many terms satisfying $\forall \varepsilon &gt;0\Rightarrow x-\varepsilon&lt;x_n$ which implies that $x$ is a limit point. I'm not entirely sure how to rule out a second limit point. Would it be sufficient to show that if $a$ and $b$ are limits of a convergent sequence $x_n$ then $a$ must equal $b$?  In the backwards direction if $x_n$ is bounded and has only one limit point then we know that there exists only one point $x$ such that we can find infinitely many terms satisfying $\forall \varepsilon &gt;0$ $\left | x_n-x\right |&lt;\varepsilon$ which implies that $x_n$ converges to $x$.   I'm quite concerned I am proving this incorrectly since I don't think I am using any facts about subsequences although they are mentioned directly in the definition for a limit point.
3370
Can anyone prove this question? I tried but I didn't get any I idea, so I hope someone can solve it.     Let $X_n\leq Y_n$ for each $n\in \Bbb N$. Show that $\liminf X_n \leq \liminf Y_n$ and $\limsup X_n \leq \limsup Y_n$.   Please prove this question - thanks.  The definition I have:     Let $X_n$ be a sequence in real number and let $$E=\{x\in \Bbb R^\sharp:(X_{n_k}) \rightarrow x \text{ for some subsequence }(X_{n_k})\text{ of }(X_n)\}$$ for all $n \in \Bbb N$ and $k$ from $1$ to $\infty$. Then by definition $\lim\sup X_n = \sup E$ and $\lim\inf X_n = \inf E$.
48176
I'm having trouble with this homework question:  "Let there be a sequence $\left(b_{n}\right)_{n=1}^{\infty}  $ such that $\lim_{n\rightarrow\infty}\left(b_{n}\right)=1$. Let there also be some bounded sequence  $\left(a_{n}\right)_{n=1}^{\infty}$  Prove that the following statement is correct: $\limsup\left(b_{n}a_{n}\right)=\limsup\left(a_{n}\right)  $"  Alright so I sat a good two hours on this question with no major breakthroughs as of yet. I have some semi-proofs all of which use some arbitrary assumptions which I can't make.    Any hints?  Thank you!
8956
a) Let $a&gt;0$ and the sequence $x_n$ fulfills $x_1&gt;0$ and $x_{n+1}=\frac{1}{2}(x_n + \frac{a}{x_n})$ for $n \in \mathbb N$. Show that $x_n \rightarrow \sqrt a$ when $n\rightarrow \infty$. I have done it in two ways, but I guess I'm not allowed to use the first one and the second one is incomplete. Can someone please help me?  We already know $x_n \rightarrow \sqrt a$, so we do another step of the iteration and see that $x_{n+1} = \sqrt a$.  Using limit, $x_n \rightarrow x, x_{n+1} \rightarrow x$ (this is the part I think it's incomplete, don't I have to show $x_{n+1} \rightarrow x$, how?), we have that   $$x = \frac x 2 (1 + \frac a {x^2}) \Rightarrow 1 = a/x^2 \Rightarrow x = \sqrt a$$ b) Let the sequence $x_n$ be defined as $x_{n+1}= 1 + \frac 1 {x_n} (n \in \mathbb N), x_1=1$. Show that it converges and calculate its limit. &quot;Tip: Show that sequences $x_{2n}$ and $x_{2n+1}$ monotone convergent to the limit.&quot; I didn't understand the tip, how can this help me? Does it make a difference if the number is odd or even? Thanks in advance!
15989
This is half of Theorem 3.19 from Baby Rudin. Rudin claims the proof is trivial. What I've come up with so far doesn't seem trivial, however, and is probably also wrong (my problem with it is pointed out below). This makes me wonder whether I'm overlooking some useful fact and/or using an unprofitable approach.  Theorem. If $s_n \leq t_n$ for $n \geq N$, where $N$ is fixed, then  $$ \liminf_{n \rightarrow \infty} s_n \leq \liminf_{n \rightarrow \infty} t_n. $$  Proof. Suppose that $s_n \leq t_n$ if $n \geq N$, but that  $$ \liminf_{n \rightarrow \infty} s_n &gt; \liminf_{n \rightarrow \infty} t_n. $$  Let $E_s$ denote the set of all subsequential limits of $\{s_n\}$, and let $E_t$ denote the set of all subsequential limits of $\{t_n\}$. Then  $$ \inf E_s &gt; \inf E_t. $$  This implies that there exists some $x \in E_t$ such that $\inf E_s &gt; x &gt; \inf E_t$, since otherwise $\inf E_t$ would not be the greatest lower bound of $E_t$. Hence some subsequence of $\{t_n\}$, say $\{t_{n_i}\}$, converges to $x &lt; \inf E_s$.  Lemma. (from Rudin) If $x &lt; \liminf_{n \rightarrow \infty} s_n$, then there exists an integer $N$ such that if $n \geq N$, then $s_n &gt; x$.   By the lemma, there exists an integer $N_0$ such that if $n \geq N_0$, then $s_n &gt; x$.   Now, let $\epsilon = \inf_{n \geq N_0} \{s_n - x\}$. This is where I think my proof breaks down. Can't $\epsilon$ be zero? Then, since $\{t_{n_i}\}$ converges to $x$, there exists an integer $N_1$ such that if $n_i \geq N_1$, then $|t_{n_i} - x| &lt; \epsilon$. But this means that, if $n_i \geq \max\{N, N_0, N_1\}$, we have both  $$ s_{n_i} &gt; t_{n_i}, $$  as well as $s_{n_i} \geq t_{n_i}$, a contradiction.