Linked Questions

0
votes
3answers
130 views

Proving a sequence is a cauchy [duplicate]

Let's suppose that $a_n$ satisfy $|a_{n+1}-a_n|<2^{-n}$ for all $n$ . How can I prove that $a_n$ is a cauchy sequence?
5
votes
3answers
2k views

A sequence of real numbers such that $\lim_{n\to+\infty}|x_n-x_{n+1}|=0$ but it is not Cauchy

Give an example of a sequence $(x_n)$ of real numbers, where $\displaystyle\lim_{n\to+\infty}|x_n-x_{n+1}|=0$, but $(x_n)$ is not a Cauchy sequence
3
votes
3answers
389 views

$\{a_n\}$ is a Cauchy sequence, if $a_{n+2} = \frac{a_n + a_{n+1}}{2}$

Suppose that the sequence $\{a_n\}$ satisies the relation $$ a_{n+2} = \frac{a_n + a_{n+1}}{2}, $$ for all $n \in \mathbb{N}_{+}$ Prove that $\{a_n\}$ is a Cauchy sequence ...
8
votes
1answer
832 views

If $\lim_{n\to\infty} (a_{n+1}-a_n)=0$ and $|a_{n+2}-a_n|<\frac{1}{2^n}$ then $(a_n)$ converges

Let $(a_n)$ be a sequence such that $\lim_{n\to\infty} (a_{n+1}-a_n)=0$ and $|a_{n+2}-a_n|<\frac{1}{2^n}$ for all $n$. I have to decide whether or not $(a_n)$ converges. My attempt: I think it ...
2
votes
2answers
206 views

What makes a sufficient proof?

This question is related to the question posted here. Would a shorter proof to those in the answers, such as: Take the subsequence $\{a_m\}$ of $\{a_n\}$ where $m > 0$. By induction on $m$ $$\...
1
vote
2answers
201 views

Showing that $\frac{\sin(a_{n-1}) + 1}{2}$ is a Cauchy sequence

In my homework set, I have the following question: Show that $$ a_n = \frac{\sin(a_{n-1}) + 1}{2}, \quad a_1=0 $$ satisfies the definition of Cauchy sequence. As we went over the concept ...
2
votes
1answer
338 views

Arithmetic mean of a sequence converges then the sequence also converges.

Let $\{a_n\}_{n\geq0}$ is a sequence of complex numbers such that $\frac{a_0+\cdots+a_n}{n+1}$ converges to $a$. Suppose also that $\exists M>0$ such that $n|a_{n}-a_{n-1}|<M$ for all $n\geq 1$. ...
1
vote
1answer
77 views

Proving that a sequence is convergent

I have to prove that for a sequence $\{a_n\}$ with $|a_{n+1} - a_n| < 2^{-n}$ is convergent. So I thought, that if a be a series then it will convergence against $\dfrac{1}{2^n}$ and for $n>1$ ...