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Let s := [(-1)i*1/i]i. We show that s converges to 0.
Take arbitrary epsilon > 0. We have to find some n in N such that forall i >= n: |(-1)i*1/i - 0| < epsilon which can be simplified to
forall i >= n: 1/i < epsilon .Take n := such n in N: 1/ epsilon < n. Because epsilon > 0, we know 1/ epsilon in R>0. Because N is unbounded (i.e., forall r in R: exists n in N: r < n), we know 1/ epsilon < n and thus n > 0. Take arbitrary i >= n. We know i > 0 and
1/i <= 1/n < 1/1/ epsilon = epsilon .