Let $\{ a_{n}\}$ be a sequence of positive real numbers such that $a_{n}=\sqrt{a_{n-1}a_{n-2}}$ for $n≥3$, then $\{ a_{n}\}$  converges to $(a_{1}a_{2}^{2})^{\frac{1}{3}}$.  My attempt:-  I multiplied the all new terms and simplify the terms based on the recursive relation   $a_{n}.a_{n-1}...a_{2}.a_{1}=\sqrt{a_{n-1}.a_{n-2}}.\sqrt{a_{n-2}.a_{n-3}}...a_{2}.a_{1}$  Cancel the like terms,  I get $a_{n}\sqrt{a_{n-1}}=a_{2}\sqrt{a_{1}}$.  Limit , I am getting the result. How is my steps? Does it have any mistakes? How to prove the existence?. I am not able to judge whether it is monotonically decreasing/ increasing. How to prove the sequence is bounded?
33499
We're given a sequence defined by the recursive relation: $$a_n = \sqrt{a_{n-1}a_{n-2}}$$  $a_1$ and $a_2$ are positive constants. We have to show the following:   The sequences $\{ b_n \} = \{ a_{2n-1} \}$ and $\{ c_n \} = \{ a_{2n} \} $ are monotonic, and if one is increasing, the other is decreasing. The limit of the sequence $\{ a_n \} $ is $ \left(a_1a_2^2\right)^{\frac13} $     Now, I have proved the first part. Besides that, I have also proved a few other things:  If $a_1 &gt; a_2$, then:  a. $ \{ b_n \} $ decreases, and $ \{ c_n \} $ increases.  b. $ c_n &lt; b_n $  If $a_1 &lt; a_2$, we just flip $\{ b_n \}$ and $\{c_n\}$  Besides, I have also shown that both the sequences : $\{b_n\}$ and $\{ c_n\}$ have the same limit. What I don't know, is how to evaluate the limit.
80485
$(a_n)$ is a bounded sequence with the following condition  $a_{n+1}\geq a_n-\frac{1}{2^n}$   The sequence converges, but how do we show it?
7641
The problem states:  a sequence of real numbers $a_0, a_1, \dots $ is defined as follows. $a_0$ is an arbitrary real number and for $n \ge 0, a_{n +1} = \lfloor a_n \rfloor \{ a_n \}$ (where $\{x\} = x - \lfloor x \rfloor$).  Prove that for $a_n = a_{n+2}$ for $n$ large enough.  my attempt:  I notice that this is a decreasing sequence because $$\frac{a_{n+2}}{a_{n+1}} = \frac{\lfloor a_{n+1} \rfloor \{ a_{n+1} \}}{ a_{n+1}} &lt; 1$$  since $\{ a_{n+1} \} \le 1$ (we can exclude the case where it's equal to one since then it becomes the null sequence that satisfies the thesis).  Moreover this sequence is bounded between $0$ and $a_0$ so by the monotone convergence theorem it must have a limit.  Taking the limit I obtain $$\ell = \lim \lfloor a_n \rfloor (\ell - \lim \lfloor  a_n \rfloor) \iff \ell = - \lim \lfloor a_n \rfloor^2 / (1 - \lim \lfloor  a_n \rfloor) $$  And so if I prove that there exists $n$ s.t. $ \lfloor a_n \rfloor &lt; 1$ I am done. Because then the limit could only be $0$ and this implies that there exist infinite $a_n = a_{n+2} = 0$.  But I can't seem to prove it. Is my reasoning up to now ok? How could I proceed?  Edit:  Add original text of the problem.
80659
Let $(a_k)$ be a sequence of real numbers and let $b_k=\frac{a_1+a_2+\dots a_k}{k}$ for each $k\in \mathbb{N}$.  Prove that if $(a_k)$ converges to $\alpha\in \mathbb{R}$, then the sequence $(b_k)$ also converges to $\alpha$.  My answer: Let $\epsilon &gt; 0$. Since ($a_k$) converges to $\alpha$, then there exists a $N_{\epsilon1}\in\mathbb{N}$ such that $|a_k - \alpha | &lt; \frac{\epsilon}{2}$ for all $k \geq N_{\epsilon1}$. Notice that $\frac{a_1+...+a_n - n\alpha}{k}$ converges to $0$ as $k$ approaches infinity. Thus, there exists $N_{\epsilon2}\in\mathbb{N}$ such that $|\frac{a_1+...+a_n - n\alpha}{k}|&lt;\frac{\epsilon}{2}$ for all $k&gt;N_{\epsilon2}$. Now, let $N= $max$ (N_{\epsilon1},N_{\epsilon2})$. Then for any $k&gt;N$ we get,  $$\begin{array}{rcll} |b_k-\alpha| &amp;=&amp; |b_k - \alpha*1|&amp;\\ &amp;=&amp; |\frac{a_1+...+a_k}{k} - \frac{k\alpha}{k}|&amp;\\ &amp;=&amp; |\frac{a_1+...+a_n}{k} + \frac{[a(n+1)+...+a_k]}{k} - \frac{(k-n)\alpha}{k} - \frac{n\alpha}{k}|&amp; \text{(notice that $N_{\epsilon1}\leq N&lt;k$)}\\ &amp;=&amp; |\frac{a_1+...+a_n - n\alpha}{k} + \frac{[a(n+1)+...+a_k - (k-n)\alpha]}{k}|&amp;\\ &amp;=&amp; |\frac{a_1+...+a_n - n\alpha}{k} + \frac{[(a(n+1) -\alpha) + ... + (a_k - \alpha)]}{k}|&amp;\\ &amp;=&amp; |\frac{a_1+...+a_n - n\alpha}{k} + \frac{(a(n+1) -\alpha)}{k} + ... + \frac{(a_k - \alpha)}{k}|&amp;\\ &amp;\leq&amp; |\frac{a_1+...+a_n - n\alpha}{k}| + |\frac{(a(n+1) - \alpha)}{k}| + ... + |\frac{(a_k - \alpha)}{k}|&amp; \text{(by the triangle inequality)}\\ &amp;=&amp;|\frac{a_1+...+a_n - n\alpha}{k}| + \frac{|(n+1) - \alpha|}{k} + ... + \frac{|a_k - \alpha|}{k}&amp;\\ &amp;&lt;&amp; \frac{\epsilon}{2} + [\frac{\epsilon}{2}/k + ... + \frac{\epsilon}{2}/k]&amp; \\ &amp;=&amp; \frac{\epsilon}{2} + \frac{\epsilon}{2}*\frac{(k-n)}{k}&amp;\\ &amp;&lt;&amp; \frac{\epsilon}{2} + \frac{\epsilon}{2}&amp; \text{(since $k-n&lt;k$)}\\ &amp;=&amp; \epsilon&amp; \end{array}$$ And since we found such an $N$ for arbitrary $\alpha$, then $(b_k)$ converges to $\alpha$ by definition.  I'm not sure i kept all my variables straightened out. How does this look and if you have a better idea let me know!
63598
If $a_n$ is a sequence such that   $$a_1 \leq a_2 \leq a_3 \leq \dotsb$$   and has the property that $a_{n+1}-a_n \to 0$, then can we conclude that $a_n$ is convergent?   I know that without the condition that the sequence is increasing, this is not true, as we could consider the sequence given in  to a similar question that does not require the sequence to be increasing.  $$0, 1, \frac12, 0, \frac13, \frac23, 1, \frac34, \frac12, \frac14, 0, \frac15, \frac25, \frac35, \frac45, 1, \dotsc$$  This oscillates between $0$ and $1$, while the difference of consecutive terms approaches $0$ since the difference is always of the form $\pm\frac1m$ and $m$ increases the further we go in this sequence.    So how can we use the condition that $a_n$ is increasing to show that $a_n$ must converge?  Or is this still not sufficient?
22387
Show that the sequence $a_1=1$, $a_2=2$, $a_{n+2} = (a_{n+1}+a_n)/2$ converges by showing it is Cauchy.   My work : Need to show that for every $\epsilon \gt 0$ there exist $N$ such that $n,m\ge N \implies  | a_n - a_m|  \lt\epsilon$. $$|a_n-a_m| = \dfrac{1}{2}|(a_{n-1} + a_{n-2}) - ( a_{m-1} + a_{m-2})|$$ I feel triangle inequality might be helpful here, but really not sure how to link it to the $\epsilon$. Appreciate any help...
104768
Show the sequence isn't bounded: $a_1=1$, $a_{n+1}=a_n+\frac 1 {a_n}$.   Proof by contradiction: Let $M&gt;0$ such that $\forall n: |a_n|&lt; M$.   Let $\epsilon &gt;0 $ and for some $n=N, \epsilon: a_N=M-\epsilon&lt;M $ pluging that in the recursion: $a_{N+1}=M-\epsilon+\frac 1 {M-\epsilon}&gt;M&gt;M-\epsilon$. Contradiction.  I wondered if I could suppose about the boundary that $\forall n: |a_n|\le M$ ? The proof would basically be the same only I could drop the epsilon.
212
Given a real positive sequence $\{a_n\}$ such that $\sum_{i=1}^n a_i$ converges.  Prove that there exists a real sequence $\{c_n\}$ monotonically increasing to $\infty$ such that $\sum_{i=1}^\infty a_nc_n$ converges.  What I have done:  I have written $$\sum_{i=1}^{N+1}a_ic_i=c_{N+1}\sum_{i=1}^{N+1}a_i-\sum_{i=1}^N(c_{n+1}-c_n)S_n$$  and then try to prove the convergence by the Cauchy criterion. However my evaluation is still not good enough. I'm really stuck.
30839
Let $a:\mathbb{N}\to\mathbb{R}^+$ be a sequence such that $a_{m+n}\leq a_ma_n$. Prove that $\sqrt[n]a_n$ converges and $$\lim_{n\to\infty}\sqrt[n]a_n=\inf_{n\in\mathbb{N}}\sqrt[n]a_n.$$  So far all I have is $\sqrt[n]a_n\leq a_1$ (from $a_{m+n}\leq a_ma_n$ by induction). If it is true that $\sqrt[n]a_n$ monotone decreasing, we are done but I couldn't show it. Should I do something with $\liminf$ and $\limsup$?
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?
50124
Prove that the sequence $\sqrt{2} , \sqrt{2\sqrt{2}},  \sqrt{2\sqrt{2\sqrt{2}}} \ $ converges to $2$.   My attempt  I proved that the sequence is increasing and bounded by $2$, can anyone help me show that the sequence converges to $2$? Thanks for your help.
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.
4506
Prove that $(a_{n})_{n \in \mathbb{N}}$ is convergent with $a_{0}=1$,   $a_{n+1}= \sqrt{2a_{n}}$ and calculate its limit.      You got the info that the square-root-function is strictly monotonic   increasing and also continuous.   I'm very confused how this could be solved because there isn't really a sequence given here. It seems to be a recursive one and I also don't know how to form $a_{n+1}$.  Could I form it to $a_{n}+ a_{1}$ ?(I don't care if it's useful here but I'd like to know if this would be correct at first.)    I have started by forming to $a_{n}$:  $$2a_{n} = (a_{n+1})^{2}$$  $$a_{n}=\frac{(a_{n+1})^{2}}{2}$$  Then I also formed to $a_{1}:$  $$a_{n}+ a_{1} = \sqrt{2a_{n}}$$  $$a_{1} = \sqrt{2a_{n}} - a_{n}$$  But it seems like this won't lead to anything.. Also I don't know what we can do with the given info, $a_{0} = 1$, the continuity and monotonic increasing...
93101
Find the limit of the sequence {$a_{n}$}, given by$$ a_{1}=0,a_{2}=\dfrac {1}{2},a_{n+1}=\dfrac {1}{3}(1+a_{n}+a^{3}_{n-1}), \ for \ n \ &gt; \ 1$$  My try:  $ a_{1}=0,a_{2}=\dfrac {1}{2},a_{3}=\dfrac {1}{2},a_{4}=0.54$ that is the sequence is incresing and each term is positive. Let the limit of the sequence be $x$. Then $ \lim _{n\rightarrow \infty }a_{n+1}=\lim _{n\rightarrow \infty }a_{n}=x$ $$ \lim _{n\rightarrow \infty }a_{n+1}= \lim _{n\rightarrow \infty }1+a_{n}+a^{3}_{n-1}$$  $\Rightarrow x=\dfrac {1}{3}( 1+x+x^3)$  $\Rightarrow x^3-2x+1=0$  and this equation has three roots $x=\dfrac {-1\pm \sqrt {5}}{2},1$   So the limit of the sequence is $\dfrac {-1 + \sqrt {5}}{2}$.  how can i say that the limit is $\dfrac {-1 + \sqrt {5}}{2}$?
2555
Let $a_1$ and $b_1$ be any two positive numbers, and define $\{ a_n\}$ and $\{ b_n\}$ by   $$a_n = \frac{2a_{n-1}b_{n-1}}{a_{n-1}+b_{n-1}},$$  $$b_n = \sqrt{a_{n-1}b_{n-1} }.$$  Prove that the sequences $\{a_n\}$ and $\{b_n\}$ converge and have the same limit.   Source: Problem Solving Through Problems by Loren C. Larson.   Hint:      Use the squeeze principle.
23637
I am trying to find out if the following sequence $(a_n)_{n\geq 1}$ will either converge or diverge. $$ a_n = (-1)^n + \frac1n $$  If the limit converges, I need to find its limit.   I tried plugging in various numbers for n besides 1, but I am not entirely sure if it really does converge.
61090
Let $(a_n)_{n=0}^{\infty}$  be a sequence of real numbers satisfying $\sum_{1}^{\infty} |a_n - a_{n-1}| &lt; \infty$   Which of the following conclusions must then be true?  The series $\sum a_nx^n$ converges   Nowhere on $ \mathbb{R}$ Everywhere on $  \mathbb{R}$ On some interval containing $(-1,1)$ Only on $(-1,1)$   What I have tried is Make an inequality with $|a -b| \leq |a| +|b|$ .. but i cant proceed further .. Any sort of help will be deeply appreciable .