## . Show that $X_n \to 0$ in probability under given condition.

Let $$k > 0$$. Suppose that

$$\forall \epsilon > 0: \exists N: \forall n \geq N: P(|X_n| \geq \epsilon) \leq \epsilon k$$

Show that $$X_n \xrightarrow{P}{ 0}$$.

Attempt:

We have to show:

$$\forall \tilde{\epsilon} > 0: \lim_n P(|X_n| \geq \tilde{\epsilon}) = 0$$

This is equivalent with showing that

$$\forall \tilde{\epsilon} > 0: \forall \epsilon \in ]0, \tilde{\epsilon}[: \exists N: \forall n \geq N : P(|X_n| \geq \tilde{\epsilon}) \leq \epsilon k$$

So, let’s show this.

Let $$\tilde{\epsilon} > 0$$ and take $$\epsilon \in ]0 , \tilde{\epsilon}[$$. Choose $$N$$ such that $$P(|X_n| \geq \epsilon) \leq \epsilon k$$ whenever $$n \geq N$$. Then, if $$n \geq N$$, we have

$$P(|X_n| \geq \tilde{\epsilon}) \leq P(|X_n| \geq \epsilon) \leq \epsilon k$$

and the claim follows.

Is this correct?

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## Let $X$ be a complete metric space. Suppose $\{x_n\}$ is a sequence in $X$…

Let $$X$$ be a complete metric space. Suppose $$\{x_n\}$$ is a sequence in $$X$$ with the property that $$|x_n – x_m| < \frac{1}{n}+\frac{1}{m}$$ for all $$n, m \in \mathbb{N}$$. Show that $$\{x_n\}$$ is convergent.

I’m struggling to find where to start for this problem and would appreciate some help.

## How to prove x_{n} diverges?

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## Let $z_n = x_n + y_n$, with $(x_n)$ and $(y_n)$ strictly increasing. Prove that if $(z_n)$ is bounded above, then so are $(x_n)$ and $(y_n)$.

Let $$(x_n)$$ and $$(y_n)$$ be strictly increasing sequences, and let $$(z_n)$$ be a sequence defined by $$z_n = x_n + y_n$$ for all $$n \in \mathbb{N}$$.

Prove that if $$(z_n)$$ is bounded above, then so are $$(x_n)$$ and $$(y_n)$$.

I do not know where to start with this problem. I know that $$(z_n)$$ being bounded above means there exists some $$A \in \mathbb{N}$$ such that $$z_n < A$$ for all $$n \in \mathbb{N}$$, therefore $$x_n + y_n < A$$ for all $$n \in \mathbb{N}$$. I don’t see how this helps finding some $$B \in \mathbb{N}$$ such that $$x_n < B$$ (or $$y_n < B$$).

I have also tried proving the contrapositive but it did not get me anywhere.

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## Proof verification that the sequence $x_n = \frac{1}{n}$ converges to every point of $\mathbb{R}$ on the cofinite topology

Let $$a \in \mathbb{R}$$. Then any open set $$U$$ in the cofinite topology containing $$a$$ is of the form $$U = \mathbb{R} – \{\alpha_1, \alpha_2, \cdots, \alpha_p\}$$ for some $$p \in \mathbb{N}$$ (and of course $$\alpha_i \neq a \ \forall 1 \leq i \leq p$$). Now, take $$N = c \left(\displaystyle{\frac{1}{\min \alpha_i}}\right) + 1$$, where $$c(x)$$ stands for the ceiling of $$x$$ (i.e, $$c(2.1) = 3$$). Then it’s clear that $$x_n \in U \ \forall n \geq N$$ and we’re done.

Is this alright? I actually think I could do away with all of this and just make the argument that the tail of $$x_n$$ is eventually contained in $$U$$ since $$\mathbb{R} – U$$ has only finitely many elements and $$\{x_n\}_{n \in \mathbb{N}}$$ is infinite, but I’m not sure if that’s fine too.

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## Proving $\sqrt{n}(x_n)$ converges when $x_n = \sin(x_{n-1}), x_1=1$

This is a problem that showed up on a qual exam that I have been stuck on for a while.

Let $$\begin{equation} x_n = \sin(x_{n-1}), x_1 = 1 \end{equation}$$ Prove $$\lim_{n \rightarrow \infty} \sqrt{n} x_n$$ exists and compute its value. The problem gave the following hint: Show that $$\begin{equation} \frac{1}{x^2_{n+1}} – \frac{1}{x^2_{n}} \end{equation}$$ converges to a constant.

I have shown that $$x_n$$ converges to $$0$$, but I am unsure on how to begin bounding $$\sqrt{n}x_n$$ or how the hint is helpful in this problem. I have tried using the MVT to show $$\sin(x_n) \rightarrow 0$$ faster than $$\sqrt{n} \rightarrow \infty$$, but I didn’t get far. Any help will be appreciated.

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