# Thread: Convergence of a sequence

1. ## Convergence of a sequence

Hello,

I am stuck with this question.

Suppose $\displaystyle x_n \leq 0$ for all $\displaystyle n \in \mathbb{N}$ and that $\displaystyle \lim ((-1)^n x_n)$ exists. Show that $\displaystyle (x_n)$ converges.

Ok, here's what I done so far...
If $\displaystyle \lim ((-1)^n x_n)$ exists, then $\displaystyle ((-1)^n x_n)$is a convergent. Thus, a Cauchy sequence.
So there exists a $\displaystyle M$ such that for all $\displaystyle \epsilon > 0$, if $\displaystyle n,m>M$ then $\displaystyle |(-1)^m x_m - (-1)^n x_n| < \epsilon$
So, if we assume that $\displaystyle n > M$, then
$\displaystyle |(-1)^{n+1} x_{n+1} - (-1)^n x_n| < \epsilon$
$\displaystyle \Rightarrow |(-1)^n ( -x_{n+1} - x_n)| < \epsilon$
$\displaystyle \Rightarrow |(-1)^n||( -x_{n+1} - x_n)| < \epsilon$
$\displaystyle \Rightarrow |-x_{n+1} - x_n| < \epsilon$
$\displaystyle \Rightarrow |x_{n+1} + x_n| < \epsilon$
Since, $\displaystyle x_n \leq 0$ for all $\displaystyle n \in \mathbb{N}$, then does that mean that $\displaystyle x_n = 0$ ??

2. Originally Posted by akolman
Hello, I am stuck with this question.
Suppose $\displaystyle x_n \leq 0$ for all $\displaystyle n \in \mathbb{N}$ and that $\displaystyle \lim ((-1)^n x_n)$ exists. Show that $\displaystyle (x_n)$ converges.
We know that $\displaystyle \left| {\left| {y_n } \right| - \left| L \right|} \right| \leqslant \left| {y_n - L} \right|$.
So it follows that $\displaystyle y_n \to L\, \Rightarrow \,\left| {y_n } \right| \to \left| L \right|$.