Show is a cauchy sequence. need whenever .
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Silly question, I suppose, but does in this problem? If so, why not just use the triangle inequality?
The sequence converges, and so it's trivially Cauchy. It converges to zero because each term has modulus .
true, but the nice thing about cauchy sequences is that there's no limits involved. note that:
if we choose , then for m,n > N
NOTE: if i represents a constant, rather than the square root of -1, then just choose:
Deveno you misread slightly (it was i^m not mi) but the proof amounted to the same thing. Let e>0 be given. Then choosing N >2/e completes the proof.
indeed i did! good catch.
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