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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

Takeforalli>=n: 1/i< epsilon .

1/i<= 1/n< 1/1/ epsilon = epsilon .

Author: Wolfgang Schreiner

Last Modification: December 14, 1999