Show that the sequence $(x_n)$ defined as $x_{n+1}=\frac{1}{2}(x_n+x_{n-1}), x_0=2, x_1=7$ is convergent. My attempt is to show that $(x_n)$ is Cauchy. So, i would like to have a feedback on my proof, please. There also might be a problem in a limit calculation... I have also one question in the end of the proof. First of all we got $\forall p\ge2$ that: $|x_p-x_{p-1}|=|\frac{1}{2}x_{p-1}-\frac{1}{2}x_{p-3}|=...=(\frac{1}{2})^{p-1}|x_2-x_0|=5(\frac{1}{2})^{p-1}$ Then, by definition of a Cauchy sequence, we have to show that $\forall \varepsilon&gt;0 \ \exists N \ \forall m,n\ge N$: $|x_m-x_n|&lt;\varepsilon$. But, $|x_m-x_n|\le|x_m-x_{m-1}|+|x_{m-1}-x_{m-2}|+...+|x_{n+1}-x_n|\le 5\sum_{k=n}^{m-1}(\frac{1}{2})^k=5(\frac{1}{2^{n-1}}-\frac{1}{2^{m-1}})\le 5\frac{1}{2^{n-1}}$ and so $(x_n)$ is a Cauchy sequence. To calculate it's limit we can consider the following: $x_n=(x_n-x_{n-1})+(x_{n-1}-x_{n-2})+...+(x_1-x_0)+x_0 =5\underbrace{\sum_{k=0}^{n-1}(\frac{1}{2})^k}_{\text{geometric series}}+2=5\Big(\frac{1-(1/2)^n}{1-1/2}\Big)+2$ So $\lim_{n \to \infty}x_n=12$... (Which might be false...) Now i have some questions: (i) First of all, my limit is clearly false as if i check first 5 values it is clear that the sequence might converge to $\sim5,3$. So, I would like to understand where i commited an error. (ii) Then, if my proof is correct, i doubt if it holds for real... If we take $m=n-1$ and $\varepsilon=1$, we have a counter example, probably. So, in $|x_p-x_{p-1}|\le5\Big(\frac{1}{2}\Big)^{p-1}$, the value $5$ makes me doubting... If possible, i would like an elaborate answer and feedback for these questions and this proof, please.
5433
The problem is the following: suppose $s_1&gt;s_2&gt;0$ and let $s_{n+1}=\frac{1}{2} (s_n+s_{n-1})$ for $n\geq 2$. Show that ($s_n$) converges.  Now, here is what I figured out:   $s_2&lt;s_4$: Base Case for induction that $s_{2n}$ is an increasing sequence. Assume $s_{2n-2}&lt;s_{2n}$. Induction step: $s_{2n}&lt;s_{2n+2}$ $s_1&gt;s_3$: Base Case for induction that $s_{2n-1}$ is a decreasing sequence. Assume $s_{2n-1}&lt;s_{2n-3}$. Induction step: $s_{2n+1}&lt;s_{2n-1}$.   I have proved those two. However arguing in favor of convergence has me going around in circles. Since $s_1&gt;s_2$ and (as discovered during the formulation of Base Cases) $s_3&gt;s_4$, I figured it might be a good idea ot assume that if every odd member of the original sequence ($s_n$) is greater than the following even member, then the limit would be somewhere in between, the two (odd and even) sequences won't cross. Hence the upper and lower bounds would be $s_1$ and $s_2$ respectively. Here is how I approach this:   Assume $s_{2n-1}&gt;s_{2n}$ Show that $s_{2n+1}&gt;s_{2n+2}$   The proof as I mentioned has me running in circles. Any assistance?
55799
Consider a convergent sequence $\sum\limits_{n=1}^\infty{a_n}$, with $a_n\ge 0$. Show that the series  $$\sum\limits_{n=1}^{\infty} \frac {\sqrt{{a_n}}}{n}$$ also converges.  Hint: $pq\le \frac 12(p^2+q^2)$
78041
I would like some advice on how to solve problems like the following: Let $(x_n)$ be a sequence defined by $x_1= 3$ and $x_{n+1} = \frac{1}{4-x_n}$. Prove that the sequence converges.  My strategy is to use the Monotone Convergence Theorem, but I am having trouble showing that the sequence is decreasing and bounded below. Here's my work so far:  Decreasing: The first 4 values are $3,1,1/3,3/11$, so let's assume $x_n \leq x_{n-1} \leq \ldots \leq x_1$. Want to show $x_{n+1} \leq x_n$. We have $x_{n+1} = \frac{1}{4-x_n}$. I want to have an upper bound for the RHS, but can't find one and don't really know where to go from here.  Bounded below: I wanted to show that all values are positive, but if we assume $x_n &gt; 0$, that doesn't rule out $x_{n+1}$ from being negative.
122547
Let ${x_n}$ be a sequence in $R$ such that $|x_{n+1} - x_n|&lt; \frac{1}{n^2}$ for all $n \in N$.  Show that the sequence is convergent.  If it were $|x_{n+1} - x_n|= \frac{1}{n^2}$, could take help of the fact $\sum \frac{1}{n^2}$ is convergent and of triangular inequality.  But what to do here? Please help. Thanks in advance.
55445
Let $\{a_n\}$ be defined by $a_1 =1  $ and $a_{n+1} = 1 + {\dfrac{1}{a_n}}$ with $n \in N$. Show that $a_{n+1}= 1 + {\dfrac{1}{a_n}}$ converges.  I know the limit but how can I show that is a Cauchy sequence or that this sequence converges.
8956
a) Let $a&gt;0$ and the sequence $x_n$ fulfills $x_1&gt;0$ and $x_{n+1}=\frac{1}{2}(x_n + \frac{a}{x_n})$ for $n \in \mathbb N$. Show that $x_n \rightarrow \sqrt a$ when $n\rightarrow \infty$. I have done it in two ways, but I guess I'm not allowed to use the first one and the second one is incomplete. Can someone please help me?  We already know $x_n \rightarrow \sqrt a$, so we do another step of the iteration and see that $x_{n+1} = \sqrt a$.  Using limit, $x_n \rightarrow x, x_{n+1} \rightarrow x$ (this is the part I think it's incomplete, don't I have to show $x_{n+1} \rightarrow x$, how?), we have that   $$x = \frac x 2 (1 + \frac a {x^2}) \Rightarrow 1 = a/x^2 \Rightarrow x = \sqrt a$$ b) Let the sequence $x_n$ be defined as $x_{n+1}= 1 + \frac 1 {x_n} (n \in \mathbb N), x_1=1$. Show that it converges and calculate its limit. &quot;Tip: Show that sequences $x_{2n}$ and $x_{2n+1}$ monotone convergent to the limit.&quot; I didn't understand the tip, how can this help me? Does it make a difference if the number is odd or even? Thanks in advance!
14679
Initially, I know that:   Convergent sequences are bounded. A sequence is convergent if and only if it is Cauchy.   Theorem. All Cauchy sequence are bounded.  My proof trying. Let $x_n$ be a Cauchy. Then, $x_n$ is converget by the $2$. So, by the $1$, $x_n$ is bounded. Hence, directlty, did I prove the theorem? Because $x_n$ is bounded.
57190
Let $\{x_n\}$ be a sequence such that there exists a $0 &lt; C &lt; 1$ such that $$|x_{n+1} - x_n| \le C |x_n - x_{n-1}|.$$  Prove that $\{x_n\}$ is Cauchy. Hint: You can freely use the formula (for $C \not = 0$) $$1+C+C^2+...+C^n = \frac {1-C^{n+1}}{1-C}.$$  My attempt : $\frac {|x_{n+1} -x _n|}{|x_n - x_{n-1}|}&lt;C$. Then, $\frac {|x_{n+1} -x _n|}{|x_n - x_{n-1}|}&lt;1$. So, I thought that it goes to zero as $n \to \infty$ because the difference between $x_n$ and $x_{n+1}$ is getting smaller as $n \to \infty.$ But, I don't know how to use the hint in the question, and how to write the answer properly.  Could you give some hint?
47786
Hi there I'm trying to solve an exercise which is part of my homework and I would really appreciate a hint how to solve it.  Given the sequence $a_n=1+\frac{1}{2}+...+\frac{1}{n} - \log(n)$ it asks me to prove that it is convergent and to prove that limit when $n$ approaches infinity of $a_n$ is $c$ where $c$ is between $0$ and $1$.  So far I proved its montony and that the sequence is decreasing. I could need a hint on how to bound it though. Thanks in advance.
118707
I think the right sequence is $f_n(t)=|t-\frac{1}{2}|^{(n+1)/n}$ but I can't manage to prove it's Cauchy... (when I look at graph of $f_n$s it seems obvious, but I need it formal way.
99636
Here is an exercise:  Suppose that $\{x_n\}$ is a sequence such that $\lim \limits_{n\to\infty}(x_n-x_{n-2})=0$.  Show that:   $$\lim \limits_{n\to\infty}\frac{x_n-x_{n-1}}{n}=0 $$  Thanks.
21683
Question :    For each $n \in \mathbb N$, Define : $\gamma_n=1+\frac{1}{2}+\dots+\frac{1}{n}-\int_1^n \frac{1}{x}dx$    Prove that $\{\gamma_n\}$ is convergent.      Note 1 :  I know that $\sum_1^\infty \frac{1}{k}$ diverges. I can guess that this sequence is convergent to $0$ because of the similar terms inside the integration and outside of it.  The problem is that the integration is continuous. I mean that $x$ is not just integers.  What can we do with the values left?! ( For example, $\frac{1}{2.5}$ ) Also, Notice that this problem should be solved with some elementary tools.  ( It's from a part of my book which is even before any discussion about the Fundamental theorem of calculus. )   Note 2 : I'm not good at integration and i'm learning it. Any good details can help me.    Thanks in advance.
91702
Define a sequence of real numbers recursively as follows. Let $a_1 = 1$ and $a_{n+1} = 1 + \frac{1}{1+a_n}$.  First, show the sequence is not monotonic.  Second, show that $a_n \geq 1$ for all $n$ and conclude that the sequence is super Cauchy.
59518
If I have a sequence {$a_n$} that has the property of $\lim(a_{n+1}-a_n)=0$, does that make it a Cauchy Sequence. I think it doesn't because $a_n = \sum_{k=1}^n \frac{1}{k}$ is a counter example.   However, by definition, there exists a $M$ such that if $n \geq M$ then $|a_{n+1}-a_n| &lt; \frac{\epsilon}{m-n}$  Hence, we have $|a_m - a_{m-1}|+.....+|a_{n+1}-a_n|&lt;|a_m -a_n| &lt;\epsilon$  This proof doesn't work because I cannot be sure I can find a fixed $M$ which might change according to n.  But I wonder if I have an additional condition that says $a_n$ is bounded, I think then the proof works and that I should be able to find a fixed $M$. However, I don't know how to justify this. Maybe I am wrong. Can someone kindly help me figure out this problem. Thanks
104844
Let $y_n$ be a sequence of real numbers. Prove that if $y_n=x_{n-1}+2x_{n}$ converges then $x_n$ also converges.  Let us suppose that $y_{n}$ goes to a limit $L$. Then for all $\varepsilon &gt;0$, for sufficiently large $n$, $|y_n -L| &lt; \varepsilon$ .  But $|y_n - L| &lt; 2|x_n-\frac{L}{3}|+|x_n - \frac{L}{3}|$  But then how to proceed?
33364
Suppose the terms of the sequence of real numbers $\{a_n\}$ satisfy $|a_{n+1} - a_n| \le 2^{-n}$ for all $n$. Prove that $\{a_n\}$ is Cauchy.   My Work  So by the definition of a Cauchy sequence, for all $\varepsilon &gt; 0$ $\exists N$ so that for $n,m \ge N$ we have $|a_m - a_n| \le \varepsilon$. However, questions like  make me understand that the $2^{-n}$ condition is necessary for this to be a true statement.   So I am wondering how to appeal to the Cauchy definition for this proof. Do I prove that every convergent sequence is therefore Cauchy, and then try to prove convergence?
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?