# How do you prove that a sequence diverges?

Specifically the sequence $\{(-2)^n\}$

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The first step is to write down exactly what it means for this sequence to diverge, according to your definitions of convergence/divergence; can you do that? –  Brian M. Scott Nov 9 '12 at 0:06
It zigzags depending on $n$ odd or even, going to infinity in either direction (even $n$ to $+\infty$, odd to $-\infty$. –  coffeemath Nov 9 '12 at 0:06

## 2 Answers

Suppose there exists L such that $(-2)^n$ tends to L. Therefore we know that there exists a positive integer $N$ such that $|(-2)^N-L|<0.5$ and $|(-2)^{N+1}-L|<0.5$ Using the triangle inequality we have: $|(-2)^N-(-2)^{N+1}|<=|(-2)^N-L|+|(-2)^{N+1}-L|<1$ , thus $|(-2)^N||1--2|<1$ (a contradiction)

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You’ve overlooked the possibility that the OP is supposed to consider a sequence convergent if it has a limit in the extended reals. –  Brian M. Scott Nov 9 '12 at 0:12

You have one subsequence decreasing to $-\infty$ and another increasing exponentially to $\infty$. No convergent sequence can have two distinct subsequential limits.

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