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Suppose that $a_n$ is a sequence in $\mathbb{R}^k$ and $\lim_{n\to\infty} a_n=A$ exists. Show that $|a_n|$ converges and $\lim_{n\to \infty}|a_n|=|A|$.

I can prove the following statement, as shown below, "Show that $\lim_{n\to \infty}|a_n|=|A|$."

Proof: Suppose that $a_n$ is a sequence in $\mathbb{R}^k$ and $\lim_{n\to\infty} a_n=A$ exists. By definition of the limit, given $\epsilon>0$ there exists $N$ such that $n\geq N$ implies $$|a_n-A|<\epsilon.$$

Note that by the reverse triangle inequality, $$\left||a_n|-|A|\right|\leq |a_n-A|.$$ Thus, $$\left||a_n|-|A|\right| < \epsilon.$$

But Is showing that $|a_n|$ converges the same as showing $\lim_{n\to\infty}|a_n|=|A|$?

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  • $\begingroup$ Everything looks okay. $\endgroup$ – Jacky Chong Nov 14 '16 at 22:09
  • $\begingroup$ Yes, but Is showing that $|a_n|$ converges the same as showing $\lim_{n\to\infty}|a_n|=|A|$? $\endgroup$ – Username Unknown Nov 14 '16 at 22:09
  • $\begingroup$ Showing $\lim_{n\rightarrow \infty}|a_n| = |A| $ is stronger. $\endgroup$ – Jacky Chong Nov 14 '16 at 22:10
  • $\begingroup$ It's not too hard to show a more general statement: if $g : \mathbb R^k \to \mathbb R$ is any continuous function, and $a_n \to A$, then $g(a_n) \to g(A)$. In this case, $g(x) = |x|$. $\endgroup$ – Bungo Nov 14 '16 at 22:10
  • $\begingroup$ @UsernameUnknown It's not the same as, but the difference is only in the other direction. If your job is to prove that $|a_n|$ converges and you show that $\lim_{n\to\infty} |a_n| = |A|$, then you have proven that $|a_n|$ converges. If your job had instead been to prove that $\lim_{n\to\infty} |a_n| = |A|$ and you only proved that $|a_n|$ is convergent, then you would still have some work left to do. $\endgroup$ – Erick Wong Nov 14 '16 at 22:10

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