Show if a sequence converges, then each subsucession converges to the same limit in Euclidean space Then let $(x_{n}) \in \mathbb{R^n}$ a convergent sequence and $(x_{r}) \in \mathbb{R^n}$ a sub-succession of $(x_{n})$, it should be shown that $(x_{r})$ is convergent and its limit is equal to that of the $(x_{n})$ sequence. proof: let $(x_{r}) \in \mathbb{R^n}$ a sub-sequence of $(x_ {n}) \in \mathbb{R^n}$; given an $\epsilon&gt;0$, exists $k \in \mathbb{N}$ such that $(x_{n}) \in B(a,\epsilon)$ for all $k&gt;K$ where $a \in \mathbb{R^n}$ and is the limit of the sequence, that is to say, $\lim x_{k}=a$. now let's see what $\lim x_{r}=a$ I'm stuck, could someone help me?
73157
I need some help understanding this proof:  Prove: If a sequence converges, then every subsequence converges to the same limit.  Proof:  Let $s_{n_k}$ denote a subsequence of $s_n$. Note that $n_k \geq k$ for all $k$. This easy to prove by induction: in fact, $n_1 \geq 1$ and $n_k \geq k$ implies $n_{k+1} &gt; n_k \geq k$ and hence $n_{k+1} \geq k+1$.  Let $\lim s_n = s$ and let $\epsilon &gt; 0$. There exists $N$ so that $n&gt;N$ implies $|s_n - s| &lt; \epsilon$. Now $k &gt; N \implies n_k &gt; N \implies |s_{n_k} - s| &lt; \epsilon$.   Therefore: $\lim_{k \to \infty} s_{n_k} = s$.   What is the intuition that each subsequence will converge to the same limit I do not understand the induction that claims $n_k \geq k$
108813
I have the following sequences $(a_n)$, $(b_n)=(a_{2n})$, and $(c_n) = (a_{2n-1})$.  I have to show that:   if $(b_n)$ and $(c_n)$ both convergence to $a$, then $(a_n)$ also converges to $a$; if $(a_n)$ converges then $(b_n)$ and $(c_n)$ also converge; there exist $(a_n)$ not converging but with $(b_n)$ and $(c_n)$ converging.   I am not sure how to solve this since I only know that $(b_n)$ > $(c_n)$ > $(a_n)$ on the terms how the sequences are defined.
9554
Suppose $\{a_n\}$ is a sequence such that the subsequences $\{a_{2n−1}\}$ and $\{a_{2n}\}$ converge to the same limit, say $a$. Show that $\{a_n\}$ also converges to $a$.
81275
Question:  Let ${a_n}$ be a sequence in $\mathbb{R}$ with $\lim_{n\to\infty}a_n=a$  Prove that  $\lim_{n\to\infty}a_n^2=a^2$  Attempt: Without using any properties of limits (this is a question in a section of the book that came before any list of any limit theorems)  By definition:  $|a_n^2-a^2|&lt;\epsilon$   $|a_n^2-a^2|$ $\leq$ $|(a_n+a)(a_n-a)|$ $\leq$ $|(a_n+a)| |(a_n-a)|$ $\leq$   $(|a_n | + |a|)(|a_n -a|)$ $&lt;$ $\epsilon$  And since I know that $\lim_{n\to\infty}a_n=a$, I'll subsitute that in for $|a_n|^2$ as $a^2$   $\therefore$ $a^2-|a|^2 &lt; \epsilon$ which is true since $a^2-|a|^2$ in my case.  I wonder if this is a possible solution?
85112
Is it true that every real sequence that converges to zero has the property that every rearrangement of it also converges to zero?  I have a proof in mind, but I'm not 100% sure it's correct (although I'm pretty sure), so I just want a yes/no answer.
112864
Show if $C$ is a subset of $\mathbb{R}$ which has a supremum $\alpha$, then there is a sequence $(c_n)$ from $C$ which converges to $\alpha$.  So far I have by definition of a supremum, for each $\epsilon$ > 0 there is a c $\in$ C such that $\alpha$ - $\epsilon$ &lt; c. But how do I work this into a sequence?
31906
Here is the question from my textbook,      "Suppose that the sequence ${a_n}$ converges to $a$ and that $a&gt;0$. Show that there is an index $N$ such that $a_n&gt;0$ for all indices $n \ge N$"   by the definition of convergent sequence:  because ${a_n} \to a$, for all $\epsilon &gt; 0, \exists N \in \mathbb{Z}^+$ such that for all $n&gt;N$, $|a_n-a| &lt; \epsilon$.  But then I don't know how to proceed. I can figure out the situation (even if it is an oscillating sequence around a), if ${a_n} \to a$, then there exists a positive integer $N$ such that for all $n&gt;N$, $a_n &gt; 0$.  p.s.: this is the first time I post a question. If there are any unclear/ improvement, please let me know
32999
Let $a_n$ be the sequence in $\mathbb{R}$ given by:  $$ a_1 = \sqrt{2}, \; \; a_{n+1} =\sqrt{2+ \sqrt{a_n} }. $$  Show that this sequence converges and find the limit.  I have already proven that this sequence is increasing and bounded by $2$, so the limit exists. Now, when I'm trying to find the limit, say  $L$, I get that $L$ satisfies the equation: $$L^2 -2 - \sqrt{L}=0, $$ but I don't actually know how to solve this equation, so I would like to know if there is an  easier way to find the limit, and in case I must solve the last equation, how can I solve it?  Thanks in advance!
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.
30915
I'm sure I'm having a notational misunderstanding. Anyway, suppose $(f_n)$ is a sequence of continuous functions from a metric space $X$ to $\mathbb{R}$. So, if $(f_n)$ converges uniformly to a function $f$, then  $$\lim_{n \to \infty} \inf_{x \in X} f_n(x) = \inf_{x \in X} \lim_{n \to \infty} f_n(x)$$  and  $$\lim_{n \to \infty} \sup_{x \in X} f_n(x) = \sup_{x \in X} \lim_{n \to \infty} f_n(x).$$  Can somebody explain to me why is that? Again, this might be really simple...
64951
Let $(a_n)_{n\in\mathbb{N}}$ be a sequence such that lim inf $|a_n|=0$. Prove that there is a subsequence $(a_{n_{k}})_{k\in\mathbb{N}}$ such that $\sum_{k=1}^\infty a_{n_k}$ converges.  I'm thought about showing that you can make a subsequence that is smaller then $1/n^2$. Is this the right strategy ? Any hints ? Should I use the Cauchy criterion ?
75676
Proposition: Let $(a_n)_{n=1}^\infty$ be a sequence of real numbers. Show that $\limsup_{n\to\infty} (a_n)\leq \sup(a_n)_{n=1}^\infty$.   Any hints on how to prove it will be appreciated.
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 $l$.
117760
Consider the sequence of functions  $$ f_n(x) = \frac{x^2}{n} $$ for $x \in \mathbb{R}$  I have shown that $f_n(x)$ converges pointwise towards $0$ for $n \rightarrow \infty$ but I am not sure whether my attempt for showing that it does not converge uniformly towards $0$ for $n \rightarrow \infty$ is correct. Do you mind verifying?   By negation we have that $f_n(x)$ does not converge towards $0$ for $n \rightarrow \infty$ if   $$\exists \epsilon &gt; 0 \forall n \in \mathbb{N} \exists x \in \mathbb{R} \exists n \in \mathbb{N}: n \geq N \Rightarrow |\frac{x^2}{n}| \geq \epsilon$$  Let $\epsilon = 1$. Then for all $n \in \mathbb{N}$ we can find a $x \in \mathbb{R}$ such that for $n \geq N$ that $|x^2/n| \geq 1$. Can I then just pick $x = n^{1/2}$ so we have that $$ |x^2/n| = |(n^{1/2})^2/n| = |n/n| = 1 \geq 1 $$ which means that $f_n(x)$ does not converge uniformly towards $0$ for $n \rightarrow \infty$. Is this ok?
111427
Previous related .  After I negated the definition of a convergent sequence, I ended up with the following mathematical statement:   $$\exists\ \epsilon &gt; 0,\ \forall\ N \in \mathbb R\ \exists\ \mathbb N \ni n &gt; N : |x_n - l| \ge \epsilon$$  Is this correct? I'd like clarification...  Anyway, I'm now asked to use my negation of the definition of convergence to prove that $(a_n)=((-1)^nn)$ is divergent...  Am I right in assuming it's enough to prove that there exists at least one $\varepsilon$ such that $|(-1)^nn -\mathscr l|\ge\varepsilon$? Can I set $\mathscr l$ to be any real number I please?
49440
I need a little hint in a proof on convergence of a sequence. The problem I have is: Suppose $(x_n)$ is a monotone decreasing sequence of real numbers such that $\sum x_n$ converges, prove that:  $$\lim_{n \to \infty} nx_n=0$$  My idea is: proving this fact aims to prove that given $\epsilon &gt;0$ there's some natural $n_0 \in \mathbb{N}$ such that if $n &gt;n_0$ we have $|nx_n|&lt;\epsilon$.  Now, I know that $\sum x_n$ converges, so that given $\epsilon'&gt;0$ there's some $k_0 \in \mathbb{N}$ such that if $k &gt; k_0$ we have:  $$\left|\sum_{i=1}^{k}x_i - S\right|&lt;\epsilon'$$  Where $S = \sum x_n$. Now, since $(x_n)$ is monotone decreasing we know that we must have $x_1 &gt; \cdots &gt; x_k$ so that the sum of all $x_i$ should be less or equal to $k x_k$. So I know that I have:  $$\left|\sum_{i=1}^{k}x_i - S\right|\leq\left|\sum_{i=1}^{k}x_i\right|+|S|\leq|kx_k|+|S|$$  I feel that the proof will come from this, however I'm stuck at this point. Can someone give just a little hint on how to proceed from here?  Thanks very much in advance for your help.
56635
I am not able to solve this exercise. I've tried everything.     Exercise Let be $a_n &gt; 0$ a sequence and suppose that $\sum a_n$ converges. Prove that exists a sequence $c_n &gt; 0$ such that $\lim c_n = \infty$ and $\sum c_na_n &lt; \infty$   I don't even know if I have to find an "always-valid" closed form for $c_n$ (maybe in function of $a_n$?) or prove its existence in another way.  Thanks in advance.