. 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?

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.

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.

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.