If $a_n$ is a sequence of real numbers, then I know the following result:   If    $\underset{n\to\infty}{\lim} a_{n}$ exist then $$\lim_{n\rightarrow\infty} a_{2n} \quad \text{and} \quad \lim_{n\to\infty} a_{2n-1} \quad \text{and} \quad  \lim_{n\rightarrow\infty} a_{3n}$$ exist and equal. Now what can I say about the converse? i.e.  if I know that $$\lim_{n\rightarrow\infty} a_{2n} \quad \text{and} \quad \lim_{n\to\infty} a_{2n-1} \quad \text{and} \quad  \lim_{n\rightarrow\infty} a_{3n}$$ exist, what can i say about $\underset{n\to\infty}{\lim} a_{n}$?
120800
Does $a_n$ converges if and only if $a_{2n},a_{3n},a_{2n-1}$ converge? $a_{2n}$ is a subsequence of $a_n$ and $a_{3n},a_{2n}$ are subsequences of $a_{6n}$ so they all have the same limit But what about $a_{2n-1}$ can I say it is a subsequence of $a_{2n}$ and therefore they all have the same limit?
111111
If $ a_0 ,a_1 ,\ldots,a_k  $ are real numbers such that  $$ a_0  + a_1  + \cdots + a_k  = 0$$  Find $$  \lim_{n \to \infty } (a_0 \sqrt n  + a_1 \sqrt {n + 1}  +\cdots + a_k \sqrt {n + k} )$$   I just can't see any solution, can anyone help ?
88350
$(z_n)_{n\in \mathbb{N}}$ is a complex valued sequence such that subsequences $(z_{2n})_{n\in \mathbb{N}}$ and $(z_{2n+1})_{n\in \mathbb{N}}$ converges to same limit.  How could I show that the sequence $(z_n)_{n\in \mathbb{N}}$ converges to the same limit using $\epsilon$, $N$ definition?    For every $\epsilon &gt;0$,  $|z_{2n}-l|&lt;\epsilon$ whenever $n&gt;N_1$, and   $|z_{2n+1}-l|&lt;\epsilon$ whenever $n&gt;N_2$.  We want to show that for every $\epsilon &gt;0$, there exists $N$ such that $|z_{n}-l|&lt;\epsilon$ whenever $n&gt;N$. I tried something like  $|z_{n}-l|\le|z_n-z_{2n}|+|z_{2n}-l|$ but wasn't sure what to do next
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!
76391
If $\{x_n\}$ is a sequence which satisfies $\lim_{n \rightarrow\infty} ~ x_{n+1} - x_n= c$  where $c$ is a real positive number. Then what can be said about the convergence/ divergence, boundedness/ unboundedness of $\{x_n/n\}$.  Attempt: $\lim_{n \rightarrow\infty} ~ x_{n+1} - x_n= c$ where $c &gt;0$  => $x_n$ is unbounded and divergent.  However, I am stuck on how to relate this to convergence/divergence of $x_n/n$ . Thanks for the help.
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?
13060
Let $(x_n)_{n\in \mathbb{N}}$ be a sequence of real numbers. A well known result is:     The sequence $(x_n)_{n\in \mathbb{N}}$ converges to $a\in \mathbb{R}$ if and only if every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ converges to $a\in \mathbb{R}$.   Now, I'm working on an exercise which essentialy asks to prove the following:     The sequence $(x_n)_{n\in \mathbb{N}}$ converges to $a\in \mathbb{R}$ if and only if for every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ there is a (sub)subsequence $(x_{n_{k_j}})_{j\in \mathbb{N}}$ which converges to $a$.   Now, one direction is trivial. If in fact $(x_n)_{n\in \mathbb{N}}$ converges to $a$, then by the result I've mentioned, every subsequence of $(x_{n})_{n\in \mathbb{N}}$ converges to $a$. Let then $(x_{n_k})_{k\in \mathbb{N}}$ be an arbitrary subsequence. In that case, we know that it is a subsequence of itself, and hence we've found a (sub)subsequence which converges to $a$.  The other way around seems quite complicated for me. The reason is as follows: I must assume that every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ has a (sub)subsequence which converges to $a$ and the proof that the original one converges to $a$.  My problem is exactly that "a". If it had all (sub)subsequences converging to $a$, then by the result i've mentionted, the subsequence $(x_{n_k})_{k\in \mathbb{N}}$ would converge to $a$. In that case, we would have every subsequence converging to $a$ and the result would follow.  My idea to prove the claim was then:  First remember that a subsequence is the restriction of the original one $x: \mathbb{N}\to \mathbb{R}$ to an infinite subset $\mathbb{N}'\subset \mathbb{N}$. In that case a (sub)subsequence is the restriction of $\tilde{x} = x|\mathbb{N}'$ to an infinite subset $\mathbb{N}''\subset \mathbb{N}$. But since $\tilde{x}$ is the restriction of $x$ to $\mathbb{N}'$ and since $\mathbb{N}''\subset \mathbb{N}'$, the restriction of $\tilde{x}$ to $\mathbb{N}''$ is itself the restriction of $x$ to $\mathbb{N}''$.  In other words, every (sub)subsequence can be viewed as a subsequence of the original one. Considering this, by hypothesis every subsequence has a (sub)subsequence which converges to $a$. But by this argument, this means that we can replace every subsequence with a convergent one. And then we can use the result I stated.  Is my idea correct? Is this really the way to prove the claim? My only doubt is that I need every subsequence converging to $a$ to use the result. In this case I can't see how I can argue that this implies every subsequence converges to $a$.  How can I finish this 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!!!
17298
Question: Give a sequence $(a_n)_{n=1}^\infty$ with $a_n&gt;0$ such that root test works while ratio test does not work, that is,    $$\lim_{n\rightarrow\infty}(a_n)^{\frac{1}{n}} \text{ exists}$$   while    $$\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n} \text{ does not exist}.$$   My attempt:   Define a sequence $(a_n)_{n=1}^\infty$ such that      $$ a_n =  \begin{cases}        2^{n-1} &amp; \text{if }n \text{ is odd,} \\    2^{n+1} &amp; \text{if }n \text{ is even}.    \end{cases}  $$         Odd subsequence $(a_{n_j})_{j=1}^\infty$ implies that $\lim_{j\rightarrow\infty}(a_{n_j})^{\frac{1}{n_j}} = \lim_{j\rightarrow\infty} 2^{1-\frac{1}{n_j}} = 2 $ while even subsequence $(a_{n_k})_{k=1}^\infty$ implies that $\lim_{k\rightarrow\infty}(a_{n_k})^{\frac{1}{n_k}} = \lim_{k\rightarrow\infty} 2^{1+\frac{1}{n_k}} = 2.$     It follows that $\lim_{n\rightarrow\infty}(a_n)^{\frac{1}{n}} = 2,$ that is, root test works.      However, for odd subsequence $(a_{n_j})_{j=1}^\infty,$     $$\lim_{j\rightarrow\infty}\frac{a_{n_j+1}}{a_{n_j}} = \lim_{n\rightarrow\infty} \frac{2^{n_j+1}}{2^{n_j-1}} = \lim_{n\rightarrow\infty} 4 = 4.$$     For even subsequence $(a_{n_k})_{k=1}^\infty,$     $$\lim_{k\rightarrow\infty}\frac{a_{n_k+1}}{a_{n_k}} = \lim_{k\rightarrow\infty} \frac{2^{n_k-1}}{2^{n_k+1}} = \lim_{k\rightarrow\infty} \frac{1}{4} = \frac{1}{4}.$$     Therefore, $\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}$ does not exist, that is, ratio test does not work.  Does my example work?
18212
Given a sequence $\{x_n\}_{n=1}^\infty$ of positive numbers, satisfying $\lim_{n \to \infty} x_n^{x_n} = 4$, prove $\lim_{n\to\infty}x_n=2$. My attampt: from $\lim_{n \to \infty} x_n^{x_n} = 4$ we know $\exists n_0\in\mathbb{N}:\forall n\ \geq n_0: |x_n^{x_n} -4| &lt; \epsilon$ so $|x_n-2| = \frac{|x_n^2-4|}{|x_n+2|} \leq |x_n^2-4| \leq |x_n^{x_n} -4| &lt;\epsilon$ Which completes the proof if correct. Is it?
110283
Prove that if $x &gt; 0$ and $x_n$ is a sequence with $\lim\limits_{n \to \infty} x_n = x$, then there is a real number $N$ s.t. whenever $n &gt; N$, $x_n &gt; 0$.  This is a homework question and I'm not really sure what methods I should use to prove this, can I get a push in the right direction? I am not expecting a flat out answer as it is a homework problem, but I am stumped!
64743
Let $(a_n)_n$ be a convergent sequence of integers , what can we say about $(a_n)_n$?   (I don't understand what is meant by this question)
28260
Assume that $$\lim_{n\to\infty}(a_{n+2}-a_{n})=A$$ show that  $\displaystyle \lim_{n\to\infty}\dfrac{a_{n}}{n}$and $\displaystyle\lim_{n\to\infty}\dfrac{a_{n+1}-a_{n}}{n}$ exist and find these limits.  maybe this problem have some methods, Thank you.  since  $$\lim_{n\to\infty}(a_{n+2}-a_{n})=A$$ so there exists $N$, and for $\forall \varepsilon&gt;0$, such $$|a_{n+2}-a_{n}-A|\le\varepsilon$$
53026
$\{a_n\}$ sequence $a_1=\sqrt{6}$ for $n \geq 1$ and $a_{n+1}=\sqrt{6+a_n}$ show that it convergence and as well find $\lim \limits_{n \to \infty} a_n$  In order to show that that sequence convergence I need to show that :  $$\lim_{n \to \infty} a_n= L$$  While $L$ is finite.  Using the calculator. I assume that L=3 because :  $$\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6........}}}}=2.999 \cong  3$$  I really don't think that this method is good enough to established that $\lim \limits_{n \to \infty} a_n= 3$ since it based on intuition.  I'll be glad to hear any ideas for an established method to show this.?  Any help will be appreciated.
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.
99539
Let $(a_n)^{\infty}_{n=m}$ and $(b_n)^{\infty}_{n=m}$ be convergent sequences of real numbers. Let $x$ and $y$ be the real numbers $x:=\lim\limits_{n\to\infty}a_n$ and $y:=\lim\limits_{n\to\infty}b_n$. Show that the sequence $(\max(a_n,b_n))$ converges to $\max(x,y)$; in other words: $$\lim_{n\to\infty}\max(a_n,b_n)=\max\bigl(\lim_{n\to\infty}a_n,\lim_{n\to\infty}b_n\bigr)$$  I was not able to prove it and would appreciate your help.
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?