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Example

We show that s := [(-1)i]i is divergent. Assume that s is convergent, i.e., s converges to some limit a in R. We show a contradiction.

Let epsilon := 1/2. There exists (by definition of convergence) some n in N such that forall i >= n: |(-1)i - a| < 1/2. We thus have

|1 - a| < 1/2 /\  |-1 - a| < 1/2
We then have (using the absolute value laws)
1 = 1/2+1/2 > |1 - a| + |-1 - a|
= |1 - a| + |1 + a| >= |(1 - a) + (1 + a)| = 2

which represents a contradiction.


Author: Wolfgang Schreiner
Last Modification: December 14, 1999

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