limit inf /sup if $x_n\leq y_n$

Question

I have just 2 problems :

1) Find the $\limsup x_n$ and $\liminf x_n$ where $x_n= e^{-n}$.

2) Let $x_n\leq y_n$ for every $n\in\mathbb{N}$ . Show that $\liminf x_n\leq\liminf y_n$ and $\limsup x_n\leq\limsup y_n$

thanks..

Answer 1

I will assume that we work with the following definitions: $$ \limsup x_n = \lim_{n\to\infty} \sup\{x_k; k\ge n\}\\ \liminf x_n = \lim_{n\to\infty} \inf\{x_k; k\ge n\} $$ (There are several equivalent definitions of limit superior and limit inferior; different authors prefer different definitions.)
We know that the above limits exists (but may be infinite) for any real sequence $(x_n)$.

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The sequence $x_n=e^{-n}$ converges to $0$. If a sequence has a limit, then both $\limsup$ and $\liminf$ of this sequence are equal to this limit.

If a sequence $x_n$ converges to limit $L$ then there is an $n_0$ such that $|x_n-L|<\varepsilon$ for $n\ge n_0$, which is the same as saying that $$L-\varepsilon < x_n < L+\varepsilon \qquad \text{for }n\ge n_0.$$ This implies that for $n\ge n_0$ we have $\sup\{x_k; k\ge n\}\le L+\varepsilon$ and $\inf\{x_k; k\ge n\}\ge L-\varepsilon$; and thus $$\newcommand{\ve}{\varepsilon}L-\ve\le \liminf x_n \le \limsup x_n \le L+\ve.$$ These inequalities are true for any $\ve>0$, therefore we get $L=\liminf x_n=\limsup x_n$.

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Now if we assume that $x_n\le y_n$ for each $n$, then we also have $$\sup\{x_k; k\ge n\} \le \sup\{y_k; k\ge n\}$$ for each $n$. Taking limit $n\to\infty$ we get $$\limsup x_n \le \limsup y_n.$$