Specifically the sequence $\{(-2)^n\}$
How do you prove that a sequence diverges?
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$\begingroup$
sequences-and-series
proof-writing
divergent-series
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0It zigzags depending on $n$ odd or even, going to infinity in either direction (even $n$ to $+\infty$, odd to $-\infty$. – 2012-11-09
2 Answers
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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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0You’ve overlooked the possibility that the OP is supposed to consider a sequence convergent if it has a limit in the **extended** reals. – 2012-11-09
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You have one subsequence decreasing to $-\infty$ and another increasing exponentially to $\infty$. No convergent sequence can have two distinct subsequential limits.