0
$\begingroup$

Let $(a_n)$ be a sequence. Prove that $\lim_{n \to\infty} a_n=0 \Leftrightarrow \lim_{n\to\infty}|a_n|=0$

First let's prove: $$\lim_{n \to\infty} a_n=0 \Rightarrow \lim_{n\to\infty}|a_n|=0$$.

Let $\varepsilon>0$. Then $\exists n_0\in \Bbb N$ such that $|a_n-0|<\varepsilon, \forall n\geq n_0$. So $|a_n|<\varepsilon, \forall n\geq n_0$. From this I should somehow conclude that $\lim_{n\to\infty} |a_n|=0$. Can you help me out with this? How do we come to that conclusion?

  • 0
    $ε$ was arbitrary, so just let it go to $0$. – 2017-01-05
  • 1
    @Now_now_Draco_play_nicely See this http://math.stackexchange.com/questions/2071204/prove-that-lim-infty-a-n-0-then-lim-infty-a-n-0/2071211#2071211 – 2017-01-05

1 Answers 1

1

From the definition of $a_n \to 0$ you have found an $n_0$ such that $|a_n| < \varepsilon$ is true for an arbitrary $\varepsilon>0$ as long as $n \ge n_0$. The definition of $|a_n|\to 0$ requires you to show that such an $n_0$ can also be found such that for all $n \ge n_0$, you have that $\left| |a_n|-0 \right| < \varepsilon$ for any $\varepsilon>0$. Since $\varepsilon$ was arbitrary at that start, notice you can use the same $n_0$ because $\left| |a_n|-0 \right| = \left| |a_n| \right| = \left| a_n \right|$.

For the other direction, notice that if $|a_n| \to 0$, then clearly also $-|a_n| \to 0$ and: $$-|a_n| \le a_n \le |a_n|$$