Proof that {$\frac{1}{b_n}$} is bounded and $\limsup\limits_{n\rightarrow\infty}$ $\frac{1}{b_n}$=$\frac{1}{\liminf\limits_{n\rightarrow\infty}b_n}$

Question

I need help/suggestions for improvements for a task, that I try to solve as a preparation for my exam.

Let {${b_n}$}$_{n\in\mathbb{N}}$ be a bounded sequence with $b_n \neq 0$ $\forall$ n $\in\mathbb{N}$ and ${\liminf\limits_{n\rightarrow\infty}b_n}$ >0. Show that the sequence {${\frac{1}{b_n}}$}$_{n\in\mathbb{N}}$ is bounded and $$\limsup\limits_{n\rightarrow\infty}\frac{1}{b_n}=\frac{1}{\liminf\limits_{n\rightarrow\infty}b_n}$$

a) Show that the sequence is $\frac{1}{b_n}$ is bounded: Well we know ${b_n}$ is bounded with L=${\liminf\limits_{n\rightarrow\infty}b_n}$ >0. So we have $$0<L<b_n<S={\limsup\limits_{n\rightarrow\infty}b_n}$$ Since $\frac{1}{b_n}\leq \frac{1}{L}$ and $\frac{1}{b_n}\geq \frac{1}{S}$. We get $$\frac{1}{S}\leq \frac{1}{b_n}\leq \frac{1}{L}$$ Thus {${\frac{1}{b_n}}$} is bounded.

Is this proof correct?

b)Let L= ${\liminf\limits_{n\rightarrow\infty}b_n}$ and M=$\limsup\limits_{n\rightarrow\infty}\frac{1}{b_n}$.

L= ${\liminf\limits_{n\rightarrow\infty}b_n}$ $\leftrightarrow$ $\forall \epsilon >0$ $\exists n\in \mathbb{N}$ with $b_n<L + \epsilon$ (dividing by L+$\epsilon$) $\rightarrow$ $\frac{1}{L+\epsilon}$<$\frac{1}{b_n}$. Since L + $\epsilon$ is fixed, we get$\frac{1}{L+\epsilon}$$\leq$M=$\limsup\limits_{n\rightarrow\infty}\frac{1}{b_n}$.

Doing exactly the same with M=$\limsup\limits_{n\rightarrow\infty}\frac{1}{b_n}$, we get L= ${\liminf\limits_{n\rightarrow\infty}b_n}$ $\leq$$\frac{1}{M-\epsilon}$.

We let $\epsilon \rightarrow$ 0 and get $M\geq \frac{1}{L}$ and $M\leq\frac{1}{L}$. Thus $$M=\limsup\limits_{n\rightarrow\infty}\frac{1}{b_n}= \frac{1}{L}= \frac{1}{{\liminf\limits_{n\rightarrow\infty}b_n}}$$

Is this proof correct? The problem with my approach. We know L=infX $\leftrightarrow$ $\forall \epsilon$>0 $\exists$ $x\in X$ with $x<L + \epsilon$. Are we allowed to replace infX with $\liminf\limits_{n\rightarrow\infty}$X ? Also, am I allowed to divide with $L + \epsilon$ ? L could be smaller than $\epsilon$. Maybe I should write straightaway, that we let $\epsilon \rightarrow 0$....

I am thankful for any advice.

Answer 1

Just some minor points:

In a) you should write $\forall \epsilon>0,\epsilon<L$ $$0<\underbrace{\liminf_{n\rightarrow \infty}b_n}_{L>0}-\epsilon < b_n < \underbrace{\limsup\limits_{n\rightarrow\infty}b_n}_{S}+\epsilon$$ for almost all $n\in\mathbb{N}$, i.e. $\forall n\geq N$ for some $N\in\mathbb{N}$, because it could be possible that no $b_n$ satisfies $b_n \geq L$ or $b_n \leq S$ (see discussion below). Since $b_n>0$, it is clear that for the finitely many $b_n$ with $n<N$ you still have boundedness of $1/b_n$.

For b) I'm not quite sure what you do there. If you fix $n$ by saying $\exists n\in \mathbb{N}$, you can not take the limit (what limit for some fixed $n$?). I guess you need to work with subsequences here, i.e. there exists a subsequence $a_k\in\mathbb{N}$ s.t. $\forall k\in\mathbb{N}$ it follows $$b_{a_k} < L + \epsilon \\ \frac{1}{b_{a_k}} > \frac{1}{L+\epsilon} \,.$$ You will still find $b_n$ that violate this inequality, but these $b_n$ do not give any information about the $\limsup$ (since they are actually smaller) and so you can say $$\limsup_{n\rightarrow \infty} \frac{1}{b_{n}}=\limsup_{k\rightarrow \infty} \frac{1}{b_{a_k}}>\frac{1}{L+\epsilon} \,.$$ Similarly, we clearly know $b_n>L-\epsilon$ for all $n\geq N$ (this time) and so $$\frac{1}{b_n} < \frac{1}{L-\epsilon}\\ \limsup_{n\rightarrow \infty}\frac{1}{b_n} < \frac{1}{L-\epsilon}\,.$$ Since this is true $\forall \epsilon>0$, the result follows.