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