Prove that if a sequence of positive real numbers $(a_n)$ converges to 1, then $(1/a_n)$ also converges to 1

Question

Let $(a_n )$$^∞_n$$_=$$_1$ be a sequence of positive real numbers.

I'm trying to prove that if $(a_n)$$^∞_n$$_=$$_1$ converges to 1 then $(1/a_n)$$^∞_n$$_=$$_1$ converges 1.

What I have so far is basically nothing. I know that, using the definition of a limit, if $(a_n)$ converges to 1, then for each ε>0, where ε is a real number, there exists N that is an integer such that for all n≥N, $|a_n-1|<ε$. I'm not sure where to go from here, but I want to show that $|(1/a_n)-1|<ε_0$.

Edit: Thanks for all your help. What I have now is:

If $(a_n )$$^∞_n$$_=$$_1$ converges to 1 then for each $\varepsilon>0$, there exists $N>0$ such that for each $n\geq N$, $|a_n-1|>\varepsilon$. Thus,

there exists $N_1>0$ such that if $n\geq N_1$ then $|a_n-1|<\frac{1}{2}$

and there exists $N_2>0$ such that if $n\geq N_2$ then $|a_n-1<\frac{\varepsilon}{2}$ for every $\varepsilon>0$

If $|a_n-1|<\frac{1}{2}$ for all $n\geq N_1$ then,

$1=|1|=|(1-a_n)+a_n|\leq (|1-a_n|+|a_n|)=(|a_n-1|+|a_n|)<\frac{1}{2}+|a_n|$.

By subtracting, $\frac{1}{2}$ on both sides of $1<\frac{1}{2}+|a_n|$,

$\frac{1}{2}<|a_n|$ for all $n\geq N_1$ and, $\frac{1}{|a_n|}<2$

Let $n=\max{\{N_1, N_2\}}$. Thus,

$|1-\frac{1}{a_n}|=|\frac{1}{a_n}-1|=|\frac{a_n-1}{a_n}|=\frac{1}{|a_n|}|a_n-1|<2(\frac{\varepsilon}{2})=\varepsilon$

Answer 1

If $a_n\to 1$ therefore:$$\forall \epsilon>0,\exists N,\qquad n>N\to 1-\epsilon<a_n<1+\epsilon$$for small enough $\epsilon$ we obtain$$1-2\epsilon<1-\epsilon+\epsilon^2-\cdots={1\over 1+\epsilon}<{1\over a_n}<{1\over 1-\epsilon}=1+\epsilon+\epsilon^2+\cdots<1+2\epsilon$$therefore $$|{1\over a_n}-1|<2\epsilon$$which means that $$\lim_{n\to \infty} {1\over a_n}=1$$

Answer 2

As an alternative by contradiction, suppose $b_n=\frac1{a_n}$ such that exists a subsequence $b_{n_k}\to L \neq 1$

then

$$c_{n_k}=a_{n_k}b_{n_k}=1 \to L$$