# Limit of $a_n$ vs limit of $(-1)^na_n$ [closed]

Is it true that $\lim_{n\to\infty}a_n=0$ for some sequence $a_n>0$ if and only if $\lim_{n\to\infty}(-1)^na_n=0$ ?

This seems like intuitively it would be the case; but I am unsure.

• Yes its true. Just use the definition. – almagest May 14 '16 at 18:13
• – Arnaud D. Oct 21 at 15:38

a) $$\lim_{n\to\infty}a_n=a \implies \lim_{n\to\infty}|a_n|=|a|$$
b) $$\lim_{n\to\infty}|a_n|=0 \implies \lim_{n\to\infty}a_n=0$$
It follows from the definition: $$\lim_{n \to \infty}a_n = 0 \Leftrightarrow \forall \varepsilon>0,\exists\, n_0\in {N}\; | \;\forall n>n_0 \Rightarrow |a_n-0|<\varepsilon$$ Now, of course $|(-1)^na_n| = |a_n|$ and you could also note that $|a_n-0| = |a_n| = ||a_n| -0|$, then, if the above property holds for $|a_n|$, it has to be valid also for $a_n$.