limn→∞xn=0⇔∀ε>0 ∃N ∀n>N ∣xn−0∣<ε\lim\limits_{n\to\infty}x_n=0\Leftrightarrow\forall\varepsilon>0\ \exists N\ \forall n>N\ |x_n-0|<\varepsilonn→∞limxn=0⇔∀ε>0 ∃N ∀n>N ∣xn−0∣<ε
limn→∞∣xn∣=0⇔∀ε>0 ∃N ∀n>N ∣∣xn∣−0∣<ε\lim\limits_{n\to\infty}|x_n|=0\Leftrightarrow\forall\varepsilon>0\ \exists N\ \forall n>N\ ||x_n|-0|<\varepsilonn→∞lim∣xn∣=0⇔∀ε>0 ∃N ∀n>N ∣∣xn∣−0∣<ε
Since ∣xn−0∣=∣∣xn∣−0∣|x_n-0|=||x_n|-0|∣xn−0∣=∣∣xn∣−0∣, we obtain that limn→∞xn=0⇔limn→∞∣xn∣=0\lim\limits_{n\to\infty}x_n=0\Leftrightarrow\lim\limits_{n\to\infty}|x_n|=0n→∞limxn=0⇔n→∞lim∣xn∣=0
Example of a divergent sequence xnx_nxn such that ∣xn∣|x_n|∣xn∣ is a convergent sequence: xn=(−1)nx_n=(-1)^nxn=(−1)n
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments