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If $\{|a_n|\}$ is convergent, then $\{a_n\}$ is convergent.
If the limit as $n$ goes to infinity of $a_n$ exists, then the sequence converges.
The first statement is false when $a_n = (-1)^n$ for integers $n > 0$.
$\lim |(-1)^n| = \lim 1^n = 1$ because this sequence is just a list of ones.
BUT
$\lim (-1)^n = $DNE because this sequence alternates between $1$ and $-1$, and therefore never reaches a specific value. -
If $\{a_n\}$ is convergent, then $\{|a_n|\}$ is convergent.
The solutions say that this is true. It says that if $a_n \rightarrow L$ as $n \rightarrow \infty$, then $|a_n| \rightarrow |L|$ as $n \rightarrow \infty$. Why is this true (Why does the absolute value of a sequence go to the absolute value of it's limit? How could one be sure of this?)?
On my own, I frequently can't come up with proofs or counter-examples for true/false problems about the convergence or divergence of sequences and series. What strategy should I use to solve these types of problems?
Best Answer
The absolute value function is continuous. Continuity implies that $\displaystyle\left|\lim_{n\to\infty} a_n\right|=\lim_{n\to\infty} |a_n|$ if the limit on the left exists.
To prove that a function is continuous at a point $a$, one shows that no matter how small a positive number $\varepsilon$ is made, there is a positive number $\delta$ small enough so that if $|a-x|<\delta$ then $|f(a)-f(x)|<\varepsilon$, i.e. the distance between $f(x)$ and $f(a)$ can be made as small as you want by making the distance from $x$ to $a$ small enough. If you don't see how that can be done with the absolute value function, first consider how it might be done if $a>0$, then how it might be done if $a<0$, and finally if $a=0$.