No-monotone sequence

Question

A no-monotone sequence can have limit?
Can I consider as no-monotone sequence $a_n=\frac{(-1)^n}{n}$?

Answer 1

The sequence is not monotone for if $n$ is even then $n + 1$ is odd and $\frac{(- 1)^{n + 1}}{n + 1} < 0 < \frac{(- 1)^n}{n}$ and if $n$ is odd then $n + 1$ is even so that $\frac{(- 1)^n}{n} < 0 < \frac{(- 1)^{n + 1}}{n + 1}$. Further if $n$ is even we have $\left| \frac{(- 1)^n}{n} \right| = \left| \frac{1}{n} \right| = \frac{1}{n}$ and if $n$ is odd we have $\left| \frac{(- 1)^n}{n} \right| = \left| \frac{- 1}{n} \right| = \left| \frac{1}{n} \right| = \frac{1}{n}$. Now for $0 < \varepsilon$ there exists a $N \in \mathbb{N}$ such that $0 < \frac{1}{N} < \varepsilon$ hence if $n \geqslant N$ we have that $\left| \frac{(- 1)^n}{n} - 0 \right| = \left| \frac{(- 1)^n}{n} \right| = \frac{1}{n} \leqslant \frac{1}{N} < \varepsilon$ proving that $\underset{n - < \infty}{\lim} \frac{(- 1)^n}{n} = 0$. So the answer is yes.