$\alpha_n \in \mathbb{R}$ is a fixed sequence of real numbers, for which the following holds:  $$ \forall t \in \mathbb{R}, \lim_{n \rightarrow \infty}e^{i\alpha_nt} = 1 $$  Is it necessarily the case that $\lim_{n \rightarrow \infty} \alpha_n = 0$?
124018
Question : Is the following true for any real number sequence $x_n$ ? &quot;If $e^{itx_n}$ converges for every $t\in\mathbb R$, then $x_n$ converges.&quot; Note that $i^2=-1$.  Motivation : We know that the converse of the above proposition is true. However, I'm facing difficulty for the above proposition. Can anyone help?
92775
Let $\alpha$ be a positive number. Calculate $$\lim_{n\to\infty} \sum_{k=1}^{n} \Big(\frac{k}{n}\Big)^{\alpha k}$$  Edit: I have deleted my attempt, it didn't seem to lead me anywhere and I discovered a couple of mistakes. :)  Please give me a hint!
30401
I want to prove the following theorem     Let $\{a_{n}\}$ be a convergent sequence. Prove that $\lim\limits_{n\to \infty}a_{n+1}=\lim\limits_{n\to \infty}a_{n}$   Do I need to show that it is monotone increasing and then use the relation  $$0\leq \lim_{n\to \infty}a_{n+1}- \lim_{n\to \infty}a_{n}\leq 0 \;?$$  Please, I need help on this! Various proofs are welcome! Thank you very much for your time!
122062
If the sequence satisfies the property lim$_{n\to \infty}(a_n-a_{n-2})=0$, prove that lim$_{n\to \infty}\frac{a_n-a_{n-1}}{n}=0$.  I am experiencing difficulty showing this rather obvious result. By definition, we know that for any $\epsilon \gt 0$, there is some $N$ such that if $n \gt N$ then $|a_n-a_{n-2}|\lt \epsilon$ My guess is that I need to write $|a_n-a_{n-1}|$ as sums of $|a_n-a_{n-2}|$ from $n$ to $N$. However, I am stuck with forming an inequality here. I would appreciate any solutions or suggestions.
11276
I wanted to prove this claim but I'm not sure if this proof is okay. Please let me know what you think! 🌻  Let $\{a_n\}_{n=1}^{\infty}$ be a sequence, and let the set of its partial limits be $PL(a_n)=A=\{a|\exists a_{n_j}\rightarrow a\}$.  Let $a_0$ be an accumulation point of A, so, exists a sequence $\{x_n\}\subset A=PL(a_n)$ such that $x_n\rightarrow a_0$. But $\{x_n\}$ is a sequence of $a_n$'s accumulation points, (Edit) so for any $\epsilon&gt;0$ There are infinitely many terms of $a_n$ in $(x_j-\epsilon,x_j+\epsilon)$, for any $j\in \Bbb N$.   Also, for any $\epsilon&gt;0$,  exists $J\in \Bbb N$ such that for every $j\ge J$, $x_j$ is in the $\epsilon$ neighborhood of $a_0$, so and there is an infinite number of $a_n$'s terms in $(a_0-\epsilon, a_0+\epsilon)$.   It follows that (Edit) for any $\epsilon$ exists $\epsilon '$ such that $(x_j-\epsilon', x_j+\epsilon')\subseteq(a_0-\epsilon,a_0+\epsilon)$ so there are infinitely many of $a_n$'s terms in the $\epsilon$ neighborhood of $a_0$. This goes to show that $a_0$ is a limit point of $x_n$, so $a_0\in PL(a_n)$ so it is closed.
87507
Here's my question:     Let $(a_{n})$ be a sequence where $(a_{1}) &gt; 0$, defined as:   $$(a_{n+1})=\arctan*(a_{n})$$ for all $n$.   Prove that $a_{n}$ has a limit $L$ and calculate it.  Solution:   Prove that x>$arctan (x)$ for all x. Prove that above we can conclude that $a_{n+1}&lt;a_{n}$, since they are equal we multiply $a_{n}$ by $arctan(x)$. $arctan(x)=0 \space iff \space x=0$, and that doesn't happen here since $a_{1}$ is positive. Therefore, we can use the follwoing theorem:      Let r be a number such that $r \in \Bbb R$ and $0&lt;r&lt;1$, so:   $$IF \space\space \left|\frac{a_{n+1}}{a_{n}}\right| \le r \Rightarrow \lim \limits_{n\to\infty}(a_{n})=0$$    Sorry for the last theorem - I don't know how it's named in English (I'm not a native speaker).  Is my solution correct?  Thanks,  Alan
111182
I have to solve this problem.      Let $\{\lambda_n\}$ be a sequence of real number such that $\lim_{n\rightarrow\infty}\lambda_n=0$ and consider the operator $T:l^p\rightarrow l^p$, $1\leq p\leq \infty$, defined by      $$T(\{x_1,\ldots x_n,\ldots\})=\{\lambda_1 x_1,\ldots,\lambda_n x_n,\ldots\} $$   Prove that $T$ is compact.    My attempt is the following.  I have considered the operator $$ T_N:l^p\rightarrow l^p,\quad T_N(\{x_1,\ldots x_n,\ldots\})=\{\lambda_1 x_1,\ldots,\lambda_N x_N,0,0,\ldots\}. $$ $T_N$ is clearly linear and it is also compact. In fact given any bounded sequence $\{x^{(n)}\}_{n\geq 1}$ in $l^p$, the sequence $\{T_Nx^{(n)}\}_{n\geq 1}$ is bounded in $\mathbb{R}^N\subset l^p$. Then, since every bounded sequence in $\mathbb{R}^N$ has a convergent subsequence, it follows that $T_N$ is compact.   Now, I have to prove that $T$ is compact. For this I proved that $||T_N-T||\rightarrow 0$ as $N\rightarrow\infty$. In fact we have: $$ ||T_N-T||=\sup_{||x||=1}||(T_N-T)x||=\sup_{||x||=1}||T_Nx-Tx||=\sup_{||x||=1}\left(\sum_{i=N+1}^{\infty}|\lambda_i x_i|^p\right)^{\frac{1}{p}} $$ and, given $\epsilon&gt;0$, there exists $\bar n$ such that $|\lambda_n|&lt;\epsilon$, for every $n&gt;\bar n$. So if we take $N&gt;\bar n$, we get $$ \sup_{||x||=1}\left(\sum_{i=N+1}^{\infty}|\lambda_i x_i|^p\right)^{\frac{1}{p}}\leq\epsilon\sup_{||x||=1}\left(\sum_{i=N+1}^{\infty}|x_i|^p\right)^{\frac{1}{p}}\leq\epsilon $$ Therefore $||T_N-T||\rightarrow 0$ as $N\rightarrow\infty$ implies that $T$ is compact.   My questions are:  (1) is the proof of the compactness of the operator $T_N$ correct?  (2) is the conclusion correct?  Thanks
55529
I'm wondering if I have a sufficient proof of the following:   If $(a_n)$ is a sequence such that $\lim_{n \rightarrow \infty}a_n=A$, then $\lim_{n \rightarrow \infty}\frac{a_1+...+a_n}{n}=A$.   My approach:  For all $\varepsilon &gt; 0$, there exists $N$ such that for all $k&gt;N$, $|a_k -A|&lt;\varepsilon$. So we can break the limit up as follows $$\lim_{n\rightarrow \infty} \frac{a_1+...+a_k}{n} + \lim_{n \rightarrow \infty}\frac{a_{k+1}+...+a_n}{n} \overset{\epsilon \rightarrow 0}{=} 0 + \lim_{n \rightarrow \infty}\frac{nA}{n}=A$$  Is this on the right track, or am I missing something about breaking up the limit in the way I have?
56284
Past final exam question for an intro to Real Analysis course:  Let $C &gt; 0$, $0&lt;r&lt;1$ and suppose that $\forall n\in \mathbb N, |x_{n+1} - x_n| \leq Cr^n$. Please help me prove that $(x_n)$ is a Cauchy sequence. (We can assume the $\lim\limits_{n\to\infty} r^n=0,$ for $0&lt;r&lt;1$)  So, I know that a Cauchy series must satisfy $|x_{n}-x_m| &lt; \epsilon$ for any $\epsilon&gt;0, \in \mathbb R$ and for all $n,m \gt H(\epsilon) \in \mathbb N$ Note that there can't be any conditions on n and m (I saw somewhere else someone required $m&gt;n$ which you can only do if you're proving its not at Cauchy sequence, right?)  Another way of doing this is showing that it is contractive (and thus a Cauchy series) if there is a constant $a$ such that $|x_{n+1}-x_n| \leq a|x_n-x_{n-1}|$   Clearly I'm supposed to make use of $\lim\limits_{n\to\infty} r^n=0,$ for $0&lt;r&lt;1$... But I don't even know how where to start with this. As I am bumbling through this problem, a more thorough answer would be much appreciated. Thanks!
49093
Claim: If $(x_\alpha)_{\alpha\in A}$ is a collection of real numbers $x_\alpha\in [0,\infty]$ such that $\sum_{\alpha\in A}x_\alpha&lt;\infty$, then $x_\alpha=0$ for all but at most countably many $\alpha\in A$ ($A$ need not be countable).  Proof: Let $\sum_{\alpha\in A}x_\alpha=M&lt;\infty$. Consider $S_n=\{\alpha\in A \mid x_\alpha&gt;1/n\}$.  Then $M\geq\sum_{\alpha\in S_n}x_\alpha&gt;\sum_{\alpha\in S_n}1/n=\frac{N}{n}$, where $N\in\mathbb{N}\cup\{\infty\}$ is the number of elements in $S_n$.  Thus $S_n$ has at most $Mn$ elements.   Hence $\{\alpha\in A \mid x_\alpha&gt;0\}=\bigcup_{n\in\mathbb{N}}S_n$ is countable as the countable union of finite sets. $\square$  First, is my proof correct? Second, are there more concise/elegant proofs?
13129
How can I show   $\lim_{n \to \infty} a_n = 0$  $a_n = {1.3.5 ... (2n-1)\over 2.4.6...(2n)}$  I have shown that $a_n$ is monotonically decreasing. I thought to shown sequence is bounded from below then it automatically would converge and hence my question will be solve. But I'm unable to show its boundedness... Or there maybe another method to prove this. Thanks
112504
I've already proved that for any $p &gt; 0$ and for any $\alpha \in \mathbb{R}$, the sequence $\frac{n^\alpha}{(1 + p)^n}$ converges to $0$.  Now, I want to prove that $\lim_{n \to \infty} x^n = 0$ as long as $|x| &lt; 1$.  I've split it into two possible cases:  (1) $0 &lt; x &lt; 1$.  Let $p = \frac{1}{x} - 1$, so that $x = \frac{1}{1 + p}$ and $p &gt; 0$ (since $x &lt; 1$). Then we let $\alpha = 0$ to see that $$\lim_{n \to \infty} x^n = \lim_{n \to \infty} \frac{n^0}{(1 + p)^n} = 0$$  (2) $-1 &lt; x &lt; 0$.  Here we cannot use the sequence $\frac{n^\alpha}{(1 + p)^n}$ again, since if we let $p = \frac{1}{x} - 1$, then $p$ isn't necessarily positive (for example, if $x = - \frac{1}{2}$, then $p = -3 &lt; 0$).  How can I prove Case (2)?
42955
Given a sequence $a_{n}$, if I know that the sequence of even terms converges to the same limit as the subsequence of odd terms:    $$\lim_{n\rightarrow\infty} a_{2n}=\lim_{n\to\infty} a_{2n-1}=L$$  Is this sufficient to prove that the $\lim_{n\to\infty}a_{n}=L$?    If so, how can I make this more rigorous?  Is there a theorem I can state that covers this case?
116702
Is  $$\bigcup_{p&lt;\infty}\ell_p=c_0 ?$$  At least one inclusion obvious: every $p$-summable sequence converges to zero.
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
123238
Let $x_n$ be a sequence of real numbers.  Show that if $\sum_{n=1}^{\infty}x_n $ converges, then $\lim_{n\rightarrow\infty}\sum_{k=n}^{\infty}x_k=0.$  This is what I have done. We have by the Cauchy Criterion, for every $\epsilon&gt;0$ there is an $N\in\mathbb{N}$ such that $n\ge N \ \implies |\sum_{k=n}^{\infty}x_k|&lt;\epsilon.$ Taking the limit as $n \rightarrow \infty$ we get the desired result.  Are there any problems with my proof?
28033
Proof that $\lim{a_n}=L$ when $n$ goes to infinity, then $\{a_n\}$ its a Cauchy Sequence  I start with this hypothesis  $\displaystyle\lim_{n \to\infty}{a_n}=L \Leftrightarrow{\forall{\epsilon}&gt;0}$  $\exists{N}\in{\mathbb{N}}$ such that  $\forall{n}&gt;N \left |{a_n-L}\right |&lt; \epsilon$  The same its true fot any $m&gt;N,\left |{a_m-L}\right |=\left |{L-a_m}\right |&lt; \epsilon$  I will sum the inequalities but I havent any idea how do it... Help me please!!!