# Questions tagged [cauchy-sequences]

For questions relating to the properties of Cauchy sequences.

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### Proving a sequence is Cauchy given some qualities about the sequence

I've got a sequence $x_n$ such that I've proved $b\leq x_n \leq c$, and $|x_{n+1}-x_{n}|\leq \frac{4}{9}|x_n-x_{n-1}|$ However I'm not very familiar with Cauchy sequences, so I don't know how to ...
2answers
108 views

### Proving Cauchy Given Given Sequence Terms Arbitrarily Close

I need to show that $\{x_{n}\}$ is Cauchy given that there exists $0<C<1$ s.t. $|x_{n+1}-x_{n}|\leq C|x_{n}-x_{n-1}|$. Intuitively, that statement clearly implies $\{x_{n}\}$ is Cauchy, since ...
4answers
118 views

### If $(a_{2n+1})$ and $(a_{2n})$ converge to $a$ then $(a_n)$ converges to $a$.

If $(a_{2n+1})$ and $(a_{2n})$ converge to $a$ then $(a_n)$ converges to $a$. So far I realize that if $(a_{2n+1})$ and $(a_{2n})$ converge then for each $\epsilon>0$, there exists $N$ such that ...
2answers
86 views

### Question on Cauchy Sequence definition?

Kaplansky defines a Cauchy sequence if for any $\epsilon > 0$ there exists sufficiently large $i, j$ such that $D(x_i, x_j) < \epsilon$ for some sequence $\{x_n\}$ in a metric space. The ...
0answers
276 views

### Completeness of metric space induced by outer measure (similar to Nikodym metric)

Let $S_\mu$ be a semi-ring of subsets of $X$ and $\mu$ be a $\sigma$-additive measure on $S_\mu$. Let $\mu^*$ be the induced outer measure on $P(X)$. Define a relation $\sim$ on $P(X)$ by \begin{align}...
0answers
176 views

### On Cauchy Sequences

I would consider this a soft question because I am seeking some insight on how to work with Cauchy sequences by using the Cauchy criterion for convergence. To my understanding, the definition is ...
1answer
1k views

1answer
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3answers
20k views

### How to generate a Cauchy random variable

How do I calculate a Cauchy random variable and how do I calculate the probability mass function to show it is "heavy tailed"
2answers
325 views

### Cauchy complete subspace of a metric space is closed.

Is some version of the Axiom of Choice required to show that a complete subspace $A$ of a metric space $X$ is closed? By closed I mean that the set's complement is open, or equivalently that it ...
3answers
569 views

### How can you prove $\{x_n\}$ is a Cauchy sequence if $|x_{n+1} – x_n| \le r|x_n - x_{n-1}|$ for some $0<r<1$?

Let $\{x_n\}$ be a sequence and let $r$ be a number such that $0 < r < 1$. Suppose that $$|x_{n+1} – x_n| \le r|x_n - x_{n-1}|$$ for all $n>1$. Prove that $\{x_n\}$ is a Cauchy sequence. I ...
4answers
210 views

### How to determine what kind of Cauchy sequences lie in a given space?

I understand the main principles of Cauchy sequences and metric spaces, but I have a particular question about determining whether or not a space is a complete space. If a space has all cauchy ...
2answers
6k views

### A “non-trivial” example of a Cauchy sequence that does not converge?

A Cauchy sequence doesn't necessarily converge, e.g. take the sequence $(1/n)$ in the space $(0,1)$. Maybe my intuition is wrong but I tend to think of this as, "it does converge but what it ...
2answers
402 views

### Showing $\mathcal{H}$ is a hilbert space.

So this is an early exercise in Conway's A Course In Functional Analysis. I'm trying to get to grips with this upto open mapping and closed graph to see if I want to do any more functional analysis. ...
2answers
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2answers
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### An issue in constructing the real numbers using Cauchy sequences in the rationals.

I'm having trouble justifying why we can discuss Cauchy sequences before the real numbers are constructed. As we know, the definition of a Cauchy sequence (in the metric space $\mathbb{Q}$) starts ...
1answer
1k views

### Given $\Sigma a_n$ diverges show that $\Sigma \frac{a_n}{1+a_n}$ diverges. [duplicate]

Intuitively speaking, I first thought that if the series $\Sigma a_n$ is divergent then $$\lim_{n \to \infty} a_n \ne 0$$ therefore it was clear that $\Sigma \frac{a_n}{1+a_n}$ would be divergent, ...
1answer
248 views

### Given $\Sigma a_n \to \alpha$ show that $\Sigma \frac{\sqrt{a_n}}{n} \to \beta$ [duplicate]

I am trying to prove that if $\Sigma a_n$ is convergent, then $\Sigma {\sqrt {a_n} \over n}$ is also convergent. I tried to use the comparison test but $\sqrt{a_n} >a_n$ so I couldn't go that ...
1answer
589 views