1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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 $\frac{1}{a}$.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

Hint

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.