Sequence (1^n) Diverges using Subsequences | Real Analysis

We prove the sequence (-1)^n diverges by finding two subsequences of (-1)^n that converge to different limits. We previously proved (-1)^n diverges by using a contradiction argument, assuming it does converge to some real number L and showing an absurdity. However, now that we have proven a sequence converges to L if and only if all of its subsequences converge to L, we have a much quicker way of proving divergence for oscillating sequences! #RealAnalysis

Note that this technique does not work for divergent sequences that aren't oscillating. If a sequence diverges to +/-infinity, all of its subsequences do as well, and so it will not be possible to find two subsequences with different limits.

Intro to Subsequences:    • Intro to Subsequences | Real Analysis  
Proof (-1)^n Diverges:    • Proof: Sequence (-1)^n Diverges | Rea...  
Sequence Converges if and only if all its Subsequences Do:    • Sequence Converges iff Every Subseque...  

Real Analysis Playlist:    • Proof: Product of Absolute Values is ...  

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