Given a sequence $(s_n)$ in $\mathbb R$ such that $$\lim \limits_{n \to \infty}( s_{n+1}-s_n)=0,$$  I am asked to prove  $(s_n)$ converges.   I know all Cauchy sequences converge in $\mathbb R^k$. So I want to prove that $(s_n)$ is Cauchy.   I am stuck as to how to show the given sequence is a Cauchy. Thank you.
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
60248
If $ (x_{n})_{n \in \mathbb{N}} $ is a sequence in $ \mathbb{R} $ and $ \displaystyle \lim_{n \to \infty} (2 x_{n + 1} - x_{n}) = x $, then is it necessarily true that $$ \lim_{n \to \infty} x_{n} = x? $$  Could anyone kindly offer suggestions on how to approach this problem? Thanks!
88024
I am trying to understand better Cauchy sequence in the book that I am reading he is trying to provide counter argument that in general metric spaces we can find sequence that are Cauchy but aren't convergent for example $\frac{1}{n}$ is Cauchy but doesn't converge in the space T = (0,1].  I am trying to prove that the sequence is Cauchy to begin with if we consider $n \geq m &gt; 0$ Consider $|a_n - a_m| = \frac{m - n}{mn} &lt; \frac{m}{mn} = 1/n &lt; \epsilon$. So if we choose N = $\frac{1}{\epsilon}$ that should work since for $n, m &gt; N = \frac{1}{\epsilon}$   we have the following:  $|a_n - a_m| = \frac{m - n}{mn} &lt; \frac{m}{mn} = \frac{1}{n} &lt; \epsilon$.  what do you guys think any the proof is good or is there anything to make it better ? It is always good to learn from the extra bit of details as it makes my brain understand the subject more !
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.
66881
First the textbook proves that all Cauchy sequences are bounded, and so have a convergent subsequence, $\{a_{n_{k}}\}$ that converges to a limit, say $L$. Now we use this to prove that all Cauchy sequences are convergent.  So an $N_1$ exists such that $$\left|a_{n_{k}}-L\right|&lt;\frac{\epsilon}{2}$$ for all $ k &gt; N_1$, and an $N_2$ exists such that $$\left|a_m-a_n\right|&lt;\frac{\epsilon}{2}$$ for all $n,m &gt; N_2 $.  Pick any $k &gt; N_1$ such that $n_k &gt; N_2$. Then for every $n &gt; N_2$, $$\left| a_n - L \right | \leq \left| a_n - a_{n_{k}} \right| + \left| a_{n_{k}} - L \right| &lt; \epsilon/2 + \epsilon/2 = \epsilon$$.  So $$\lim_{n \rightarrow \infty}a_n = L$$  I'm fine with this proof until the last part - I'm confused as to why we can pick an abitrary $k$ like we do? Should the limit not depend only on $n$? Now it appears like it also depends on $k$ and if we pick a $k &lt; N_1$, the inequality isn't true. Can someone clarify this for me please?
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.
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...
79021
I am looking for an example of a sequence which shows that, $a_{n+1} - a_n \to 0$ as $n \to \infty$ does not imply that sequence $a_n$ converges. I have a feeling that the sequence should be an oscillatory one but I am unable to think of an example.
53269
Let $(x_n)$ be a sequence of real numbers.  Prove that if there exists $x$ such that every subsequence $(x_{n_k})$ of $(x_n)$ has a convergent (sub-)subsequence $(x_{n_{k_l}})$ to $x$, then the original sequence $(x_n)$ itself converges to $x$ .              Thanks for any help.
33814
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
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!
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
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.
5889
I will post the exercise below:      Prove that the sequence $(a_n)$ defined by $a_0 = 1$, $a_{n+1} = 1 + \frac 1{a_n}$ for $n \in \mathbb N$ is convergent in $\mathbb R$ with the Euclidean metric, and determine afterwards is limit. Can you intepret the limit geometrically (hint: Golden ratio)?   So I need to prove that the sequence is convergent in $\mathbb{R}$  with the Euclidean metric, and how do I prove that? The limit must be $1$, but how to interpret it geometrically?
52746
Does the condition $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ imply that $a_n$ convergent?  When I say convergent I mean to $L\in \mathbb{R}$ or to $\infty$ or to $-\infty$  I have a gut feeling this statement is false, just can't find a series which contradicts it.  Thanks a lot
16916
Let $\{a_n\}$ be a sequence of real numbers such that $|a_{n+1} - a_n| \le \dfrac{1}{2^n}$.   I would like to show that this sequence is Cauchy.    Letting $\epsilon &gt; 0$, I said choose $N$ such that $1/2^N \le \epsilon$. However, I'm not sure if this is the right $N$, since $m, n &gt; N$ doesn't seem to necessarily imply $|a_n - a_m| \le \epsilon$.  Can someone show me the light?
23237
Let $(a_n)_n$ be a sequence of non-negative real numbers.      Prove that $\sum\limits_{n=1}^{\infty}a_n$ converges if and only if the infinite product $\prod\limits_{n=1}^{\infty}(1+a_n)$ converges.   I don't necessarily need a full solution, just a clue on how to proceed would be appreciated.  Could any convergence test like the ratio or the root test help?