Convergence of a sequence

**akolman** · August 17th 2010, 02:59 PM

Hello,

I am stuck with this question.

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

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

**Plato** · August 17th 2010, 03:33 PM

Originally Posted by akolman

Hello, I am stuck with this question.
Suppose $x_n \leq 0$ for all $n \in \mathbb{N}$ and that $\lim ((-1)^n x_n)$ exists. Show that $(x_n)$ converges.

Come on. Think about it.
We know that $\left| {\left| {y_n } \right| - \left| L \right|} \right| \leqslant \left| {y_n - L} \right|$ .
So it follows that $y_n \to L\, \Rightarrow \,\left| {y_n } \right| \to \left| L \right|$ .

Can you apply that to this problem?

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