If $\lim_{n \to \infty} |a_n| = |L|$ then $\lim_{n \to \infty} a_n = L$

Question

This is the question that I am having a bit of trouble with and would like some help and/or advice.

Prove or disprove. Suppose we have a sequence $\{a_n\}_{n = 1}^{\infty}$. If $\lim_{n \to \infty} |a_n| = |L|$, then $\lim_{n \to \infty} a_n = L$.

My thought of approaching this question was to prove the converse of this statement, so what I really want to prove is if $\lim_{n \to \infty} a_n = L$, then $\lim_{n \to \infty} |a_n| = |L|$.

For the limit $\lim_{n \to \infty} a_n = L$, by the definition of the limit: $\forall\varepsilon > 0 \exists N\in\mathbb{N}$ such that if \begin{equation*} n > N \hspace{0.5cm} \rightarrow \hspace{0.5cm} |a_n - L| < \varepsilon \end{equation*}

But we also know that for the limit $\lim_{n \to \infty} |a_n| = |L|$, we have: $\forall\varepsilon > 0 \exists N\in\mathbb{N}$ such that if \begin{equation*} n > N \hspace{0.5cm} \rightarrow \hspace{0.5cm} ||a_n| - |L|| < \varepsilon \end{equation*}

I was wondering how to proceed from here. This is the part where I am having a bit of trouble with. Some advice would be great!

Answer 1

You can show a counterexample. Take $a_n=-1$ and $L=1$. We have $lim_{n \to \infty} |a_n| = |L|$, but $lim_{n \to \infty} a_n \neq L$.

Answer 2

I am not an expert with sequence, but for this one you can "see through" if you take $a_n = (-1)^n$. Hope this helps.

Answer 3

As, $lim_{n \to \infty} a_n = L$, then for $\varepsilon >0 $, $\exists$ a natural number $k \in \mathbb{N}$, such that $\mid a_{n} - l\mid < \varepsilon ,\forall n \geq k$.

Now, $\mid \mid a_{n} \mid- \mid l \mid \mid \leq \mid a_{n}-l\mid < \varepsilon, \forall n \geq k$, by using the property : $\mid \mid a \mid- \mid b \mid \mid \leq \mid a-b\mid$.

So, finally we get : $lim_{n \to \infty} \mid a_{n} \mid = \mid l \mid$. (Proved)

Now, time for a counterexample :

Let, $\{a_{n}\} = \{(-1)^n\}$. Now, for $\mid \{a_{n}\}\mid = \{1,1,1,...\}$ converges to 1 (as it is a constant sequence), but $\{a_{n}\}$ doesn't converge as it oscillates between two finite values : $\{-1,1\}$.