My classmate asked a question during lecture about our discussion of bounded sequences, particularly the sequence $\sin(n)$. His question was, What is $\sup(\sin(n))$?
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how to prove  $\sup \{ \sin n \mid n\in \mathbb N \} =1$
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Suppose that $(u_n)$ is unbounded above. Then, we pick any arbitrary monotone increasing subsequence $(v_n)$ of $(u_n)$. But by hypothesis, we can find a subsequence of $(v_n)$ that converges to $0$. Hence, $\lim v_n =0$. Therefore, $(u_n)$ must be bounded above.  Again, let $(u_n)$ be unbounded below. Then we pick any monotone decreasing subsequence $(w_n)$. By similar arguments, $(u_n)$ must be bounded below.   Being bounded, it has a finite $\limsup =l$ and $\liminf =m$.  Hence, there must be convergent subsequences $(a_n)$ and $(b_n)$ that converge to $l$ and $m$ respectively. But every subsequences of the above two must converge to $l$ and $m$ respectively (treating the two as convergent sequences). Again, by hypothesis, they both have subsequences that converge to $0$. Therefore, both $(a_n)$ and $(b_n)$ converge to $0$.   In conclusion, $\limsup u_n = \liminf u_n = 0 =\lim u_n$  Is this alright?
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Let $1&lt;p&lt;\infty$, $(u_n)_{n\in\mathbb{N}}\subseteq l^p(\mathbb{N})$ such that $u_n$ converges weakly to $u\in l^p(\mathbb{N})$. Prove that   (1) $(u_n)$ is bounded and (2) for all fixed component $j\in n\in\mathbb{N}$ it is $\lim_{n\to\infty} u_{n,j}=u_j$ (in lecture we called it "pointwise convergence of sequences").   My questions are: (1) is "$(u_n)$ bounded", i.e. $\|u_n\|&lt;\infty$ for all $n\in\mathbb{N}$ already satisfied because $(u_n)_{n\in\mathbb{N}}\subseteq l^p(\mathbb{N})$?   For (2) it is to show that $|u_{n,j}-u_j|\to 0,\; n\to\infty$ for fixed $j\in\mathbb{N}$. But I don't know how to do this.
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I'm trying to prove that; If any Cauchy sequence is convergent in an ordered field F, every nonempty subset of F that has an upperbound has a sup in F.  Let A be a nonempty subset of F that is not a singleton and has an upperbound in F. Let $a_0 \notin v(A)$ and $b_0 \in v(A)$. It's written in my book that for every $e \in P_F$, there exists $N \in \omega$ such that $N≧(b_0 - a_0)/e$.  I think this is not accurate since it hasn't showed that such F is Archimedean.. Is such F archimedean? Or in such a condition does there exist such N?  -Definition of a Cauchy sequence; For every $e\in P_F$, there exists $N\in \omega$ such that if $i,j≧N$, then $|x(i) - x(j)| &lt; e$. ($x:\omega \to F$ is a sequence)
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Be $ 0 &lt; q &lt; 1, (a_n)_{n \in  \mathbb{N}} $ a sequence in $\mathbb{R}$ and $ n_0  \in  \mathbb{N} $  One has: $ |a_{n+1} -a_n| \leq q |a_n-a_{n-1}|$  for all $n \geq n_o$. Show, that the sequence $(a_n)_{n \in  \mathbb{N}}$ convergates.  I tryed to verify, that its a Cauchy-sequence, but i didn't gone very far..
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I think I have the basic framework for this proof, but I am having trouble putting everything together in a convincing way.    I plan on showing that by the definition, an unbounded sequence $(a_n)$ has infinitely many $n$ such that $a_n&gt;M$ where $M&gt;0$, for some natural number $n$. I'm not sure how it is implied that there exists some $a_{n_k}$ such that $a_{n_k}&gt;M$ for some natural number $k$.  Let me know if I'm on the right track, and please help me string these ideas together!
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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.
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Suppose that $S\neq\emptyset$ is a bounded set of numbers and that $a$ is a number. Define $aS=\{ax\mid x\in S\}$.   Prove that sup $aS$ = $a$ sup $S$ if $a \geq 0$  I can intuitively see why this is true by just trying out some cases for $a$ and $S$ but I can't seem to prove why this is true.
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Let $x_n$ be a sequence of the type described above. It is not monotonic in general, so boundedness won't help. So, it seems as if I should show it's Cauchy. A wrong way to do this would be as follows (I'm on a mobile device, so I can't type absolute values. Bear with me.) $$\left|x_{n+1} - x_n\right| \leq \frac{1}{n^2}.$$ So, we have $$ \left|x_m -x_n\right| \leq \sum_{k=n}^{m} \frac{1}{k^2}$$ which is itself Cauchy, etc., etc. But, of course, I can't just use absolute values like that. One thing I have shown is that $x_n$ is bounded. Inductively, one may show $$ \limsup_{n \to \infty} x_n \leq x_k + \sum_{k=n}^{\infty} \frac{1}{k^2},$$ although I'm not sure this helps or matters at all. Thanks in advance.  Disclaimer I've noticed that asking a large number of questions in quick succession on this site is often frowned upon, especially when little or no effort has been given by the asker. However, I am preparing for a large test in a few days and will be sifting through dozens of problems. Therefore, I may post a couple a day. I will only do so when I have made some initial, meaningful progress. Thanks.
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If $x\in \Bbb R$ ; then is the sequence $\{a_n\}$ where $a_1=x$; $a_{n+1}=\cos (a_n)$ convergent?  Obviously $|a_n|\le1$  and hence $(a_n)$ is bounded.  Also $f(x)=\cos x$ is decreasing for $x&gt;0$.  But here $x\in \Bbb R$ .How to proceed here?Please help.
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The question :     Let $D$ be a nonempty subset of the reals that is bounded above. Is the supremum of $D$ a limit point of $D$?   My Reasoning: I think this is false for these two cases. Case 1:If I look at $D = \{n \in \mathbb{Z} | n \le 0\}$ the supremum is $0$. And since I need a convergent sequence $\{x_n\} \subset D/\{0\}$ the converges to $0$ for it to be a limit point I can say in this case if I look at $\epsilon = \frac{1}{2}$ for the converges of the sequence it will fail to converge and so $0$ isn't a limit point.   Case 2: Also if I look at $D = {0}$ then the supremum is $0$. And $D$ is a subset of the reals. So if I look for a sequence $\{x_n\} \subset D/\{0\}$ I can't make one because $D/\{0\}$ is the empty set.   My question is this. Since the problem asked about an arbitrary subset of the reals $D$, can I define $D$ to give a counterexample like above or have I misunderstood the question?  --Thanks in advance.
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Please help me solve this. Please try to give some details also.  Let $a_n=\sum_{k=1}^n  \frac{n}{n^2+k}\ $for $ n \in\mathbb{N}$ then the sequence $\{a_n\}$ is (choose the correct option):     Convergent   Bounded but not convergent   Diverges to infinity   Neither bounded nor diverges to infinity     My thoughts : I tried doing it by testing if it is monotonic with help of $a_{n+1} -a_n$ and checking if its +ve or -ve but, the the terms dont simplyfy by cancelling each other...
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This is a question from the book Methods of Real Analysis by R. R. Goldberg.     If $(s_n)$ is a sequence of real numbers and if $$\sigma_n=\frac{s_1+s_2+\cdots+s_n}{n}$$ then prove that: $\operatorname{{lim sup}}\sigma_n \leq \operatorname{lim sup} s_n$.   I don't have any idea how to start working on this problem. Please help. Thanks.
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Can anybody please help me to prove this:  Let $p$ be greater than or equal to $1$. Show that for the space $\ell_p=\{(u_n):\sum_{n=1}^\infty |u_n|^p&lt;\infty\}$ of all $p$-summable sequences (with norm $||u||_p=\sqrt[p]{\sum_{n=1}^\infty |u_n|^p}\ )$, there is an inner product $&lt;\_\,|\,\_&gt; $ s.t. $||u||^2=&lt;u\,|\,u&gt;$ if and only if $p=2$.
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Note: $x_n$ is a sequence which is not necessarily convergent.    The following was my attempt. Since $\lim_{n\to \infty}a_n=a$ then $\limsup_{n\to \infty}a_n=a$ . Also $\sup(a_nx_n)=\sup(a_n)\sup(x_n)$. Therefore, $\limsup_{n\to \infty}a_nx_n=\limsup_{n\to \infty}a_n\limsup_{n\to \infty}x_n=a\limsup_{n\to \infty}x_n$.    I am not sure if this is correct. So could someone please show me how it is done. Thanks
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So one of my professors proposed a problem to me and it has stumped me for some time now. Here's how it goes:  Suppose you have a sequence $a_n$ of real numbers such that $$\lim_{n\to\infty} a_{n} = 0$$ and suppose the sequence of partial sums $s_n$ is bounded. Prove that $s_n$ converges or give a counterexample.  I'm hoping to figure this out without anyone handing me the complete solution, so if someone could point me in the right direction with a hint, it would be much appreciated.
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I have lecture notes with the claim $(C_b(X), \|\cdot\|_\infty)$, the space of bounded continuous functions with the sup norm is complete.  The lecturer then proved two things, (i) that $f(x) = \lim f_n (x)$ is bounded and (ii) that $\lim f_n \in \mathbb{R}$.  I don't understand why it's not enough that $f$ is bounded. I think the limit of a sequence of continuous functions is continuous and then if $f$ is bounded, it's in $C_b(X)$. So what is this $\lim f_n \in \mathbb{R}$ about? Many thanks for your help.