2

Prove: If $(a_n)\to\infty$, then $(1/a_n)\to0$

As $a_n\to\infty$, for every $C>0$ there exists an $N$ in the natural numbers such that we have

$a_n>C$ whenever $n>N$

$\implies|a_n|>C$ whenever $n>N$

$\implies|1/a_n| < 1/C$ whenever $n>N$

If follows that if we choose $e>0$ there exists an $N_1$ such that we have

$|1/a_n| < e$ whenever $n>N_1$

How is this proof, is it acceptable do you think?

Shubham Johri
  • 17,659
James Doherty
  • 127
  • 8
  • 1
    Almost. For $\varepsilon$ choose $C:=1/\varepsilon$ – Berci Jan 10 '19 at 17:43
  • Thanks I don't quite understand are you saying that for the last bit I should say that if we choose e=1/C the definition will be satisfied? – James Doherty Jan 10 '19 at 17:50
  • Yes, I miss that part from the proof.. But not exactly.. First, $\varepsilon$ is given, and we're looking for an $N_1$, which will be chosen for $C=1/\varepsilon$ by the hypothesis. – Berci Jan 10 '19 at 17:51
  • @JamesDoherty You need to show that $\forall\epsilon>0,\exists n_0\in\Bbb N:|1/a_n|<\epsilon\forall n\ge n_0$. To prove this, take $C=1/\epsilon>0$. Then $\exists n_1\in\Bbb N:a_n>C\forall n\ge n_1\implies 1/a_n=|1/a_n|<1/C=\epsilon\forall n\ge n_0=n_1$ – Shubham Johri Jan 10 '19 at 17:56

0 Answers0