# Absolute Value Theorem for Sequence

Given an alternating sequence.

The Absolute Value Theorem states that: If the limit of the absolute value of the sequence is 0 then the limit of the original sequence is also 0.

However if, the limit of the absolute value of the sequence is not 0 (some value) then there is "no conclusion".

But wouldn't that just imply that the sequence would alternate between this value making the limit DNE and therefore the sequence divergent? Or is this only for all cases where the sequence is in fact alternating.

• Hint: Consider the following sequences: $$a_n=(-1)^n,\qquad b_n=1$$What are $\lim\limits_{n\to\infty}|a_n|$ and $\lim\limits_{n\to\infty}|b_n|$? Do $a_n$ and $b_n$ both converge and diverge together? What can you draw from this example? – Simply Beautiful Art Jun 7 '17 at 1:13
• I guess it would be divergent for alternating sequence cases only then – mathguy Jun 7 '17 at 1:13
• But can you show that it must alternate only given that $\lim\limits_{n\to\infty}|x_n|=1$? Can you show anything given only this much information? – Simply Beautiful Art Jun 7 '17 at 1:15
• So I think you've answered your own question ;) – Simply Beautiful Art Jun 7 '17 at 1:18
• (By the way, you can try writing an answer to your very own question! I'm looking forward to it $\ddot\smile$ ) – Simply Beautiful Art Jun 7 '17 at 1:24

Assume that $$\lim_{n\to+\infty}|x_n |=L>0$$

if, for enough great values of $n$, $x_n$ keeps a constant sign then the sequence $(x_n)$ will converge to $L$ or $-L$.

if its sign changes, it will diverge.

The resulting limit from using the Absolute Value Theorem must be 0 to show convergence of the original sequence.

The resulting limit can also be a non zero number which, in general means that there is no conclusion.

However, when the resulting limit is a non zero number and the original sequence is an alternating sequence, you can conclude that the sequence will alternate between this non zero number. Making the limit of the original sequence DNE which implies that the sequence is divergent.

• Rather, if there exists some $N>0$ such that $\operatorname{sgn}(x_n)=\operatorname{sgn}(x_N)$ for all $n>N$, then the sequence is convergent, else, it is divergent. – Simply Beautiful Art Jun 7 '17 at 1:46
• @SimplyBeautifulArt This is exactly what i wrote. – hamam_Abdallah Jun 7 '17 at 1:48
• @Salahamam_Fatima :P And I've already upvoted you in sight of that. I mostly wanted to point out that alternating usually implies $(-1)^n$, but such simple sign changes are not what is needed, but rather that the sign changes eventually, or never. – Simply Beautiful Art Jun 7 '17 at 1:54
• @SimplyBeautifulArt I simply thank you. – hamam_Abdallah Jun 7 '17 at 1:56

The continuity of $x \mapsto |x|$ implies that the convergence of $\langle a_j \rangle$ implies the convergence of $\langle |a_j| \rangle$, so taking the contrapositive gives the divergence of $\langle a_j \rangle$.