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?


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

  • $\begingroup$ Tips: to make MathJax works, use the dollar symbols to wrap up your math expressions. E.g. $\mathbb N$ vs \mathbb N. $\endgroup$ – 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 $\frac{1}{a}$.

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


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

  • $\begingroup$ 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? $\endgroup$ – TUC Sep 30 '18 at 12:59
  • $\begingroup$ I think that the definition of convergence is enough $\endgroup$ – giannispapav Sep 30 '18 at 13:04

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