# Prove that $b_k \to 1/a$ if $a_k \to a$ where $b_0 = 0$ and $b_k = 1/a_k$ for $k>0$

Let $$a_n$$ where $$n \in \mathbb {N}$$ be a sequence of rational numbers converging to $$a$$. Suppose $$a \neq 0$$, for $$k = 1, 2, ...$$ let $$b_k=\begin{cases} 0 & \text{if}\;a_k=0\\\\\frac{1}{a_k} &\text{if}\;a_k \neq 0\end{cases}$$ Prove that $$b_n$$ converges to $$\frac{1}{a}$$.

I was studying real analysis and got stuck on this problem. Can you help me solve this problem or give me some hints?

Thanks

edit: is it possible to solve this in terms of Cauchy Sequence?

• Tips: to make MathJax works, use the dollar symbols to wrap up your math expressions. E.g. $\mathbb N$ vs \mathbb N. – xbh Sep 30 '18 at 12:47

It is not necessary to use Cauchy sequences.

Let $$\varepsilon = \frac{|a|}{2} > 0$$ (because $$a \neq 0)$$. The sequence $$(a_k)$$ converges to $$a$$, so there exists $$N \in \mathbb{N}$$, such that for all $$k \geq N$$, $$a-\varepsilon < a_k < a +\varepsilon$$. By definition of $$\varepsilon$$, you get $$a_k \neq 0$$ for all $$k \geq N$$.

So, for all $$k \geq N$$, $$b_k = \frac{1}{a_k}$$. Taking the limit, you get immediately that $$(b_k)$$ converges to $$\frac{1}{a}$$.

• So $\lim_{n\to \infty}$ $(b_n)$ $_n \in \mathbb N$ = $\frac{1}{\lim_{n\to \infty} (a_n)}$ = $\frac{1}{a}$ ? – TUC Sep 30 '18 at 13:41

Hint

Because $$a\not=0$$ (let's say $$a>0$$), show that $$\exists n_0\in\mathbb{N}: \forall n\geq n_0\quad a_n>0$$

• I already know that $a_n$ is a Cauchy sequence. Do I also have to show that $b_n$ is a Cauchy sequence as well? – TUC Sep 30 '18 at 12:59
• I think that the definition of convergence is enough – giannispapav Sep 30 '18 at 13:04