If $\sup \{a_n\mid n\in \mathbb{N}\}=1$ then $\frac{1}{1-a_n}\to\infty$

Question

Suppose $(a_n)$ is a sequence such that $a_n<1$ for all $n$ and $s:=\sup \{a_n\mid n\in \mathbb{N}\}=1$. I want to prove that $\frac{1}{1-a_n}\to\infty$.

My initial approach: let $M>0$ and $\varepsilon=1/M$. Then, there exists $N_1\in\mathbb{N}$ such that $a_{N_1}>s-\varepsilon=1-1/M$. Then: $$\frac{1}{1-a_{N_1}}>M$$

But to prove that $\frac{1}{1-a_n}\to\infty$ I must find $N$ such that for all $n>N$ : $\frac{1}{1-a_{n}}>M$. How can I find such $N$?

Answer 1

The claim clearly isn't true, as the sequence \begin{align}a_{2n}&=1-\frac1{2n}\\ a_{2n+1}&=0\end{align} satisfies $a_n<1$ and $\sup_n a_n = 1$, but $\frac1{1-a_{2n+1}}=1$ for all $n$, so $$\liminf_{n\to\infty} \frac1{1-a_n}=1.$$ This example also shows that $\limsup_{n\to\infty}a_n=1$ isn't a sufficient condition for $\frac1{1-a_n}\to\infty$.

However, $\lim_{n\to\infty}a_n=1$ is sufficient. If $M>0$, then choose $N$ so that $n\geqslant N$ implies $$1-a_n<\frac1M.$$ Then for $n\geqslant N$, $$\frac1{1-a_n}>M, $$ so that $$\frac1{1-a_n}\stackrel{n\to\infty}\longrightarrow \infty. $$

Answer 2

It's not a correct statement. Let $a_n = \sin n$. It's obviously that $\sup a_n = 1$; but limit $\dfrac{1}{1-\sin n}$ doesn't exists.

Answer 3

The answers given here are all good, but they choose sequences that are too complicated. The sequence $a_n = \frac 1 n$ for $n \ge 1$ has $\sup _n a_n = a_1 = 1$, but $\frac 1 {1-a_n} \to 1$, so the statement is false.