# Suppose $\lim x_{n}=x$ where $x\ne0$. Prove that the sequence $(-1)^{n}x_{n}$ is not convergent.

Suppose $\lim x_{n}=x$ where $x\ne0$. Prove that the sequence $(-1)^{n}x_{n}$ is not convergent

My goal is to find two subsequences which converge to different values. Given the subsequence $y=(-1)^nx_n$ such that $n$ is odd gives the sequence $-x_n$. Now since the $\lim x_n=x$ then $\lim (-1)x_n=-x$ likewise given the subsequence $z=(-1)^nx_n$ such that $n$ is even gives the sequence $x_n$ thus $\lim x_n=x$. Therefore we have two subsequences which converge to different values so the sequence is divergent.

I feel I am missing some major key points any help would be appreciated

• What you have done is correct. The fact that $x \neq 0$ shows that the two subsequences you have chosen converge to different limits thus proving the divergence of the sequence – Erm20 Feb 13 '18 at 8:58
• It might be better to write $y_n=(-1)^{2n+1}x_{2n+1}=-x_{2n+1}$ and $z_n=(-1)^{2n}x_{2n}=x_{2n}$. Missing a major point? Mmm.. maybe that you forgot to mention that $x\neq0$ implies that $x\neq-x$. But that's trivial of course, so I don't see you missing any major point. – drhab Feb 13 '18 at 8:58
• "Therefore we have two subsequences which converge to different values so the sequence is divergent. " Be careful with this language, not convergent is not the same as divergent. – Tony S.F. Feb 13 '18 at 10:24