Show is a cauchy sequence. need whenever .

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- October 22nd 2011, 11:46 AMDukeshowing its a cauchy sequence
Show is a cauchy sequence. need whenever .

- October 22nd 2011, 05:56 PMAckbeetRe: showing its a cauchy sequence
Silly question, I suppose, but does in this problem? If so, why not just use the triangle inequality?

- October 22nd 2011, 05:57 PMDrexel28Re: showing its a cauchy sequence
The sequence converges, and so it's trivially Cauchy. It converges to zero because each term has modulus .

- October 23rd 2011, 12:01 AMDevenoRe: showing its a cauchy sequence
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:

- October 23rd 2011, 03:23 AMDukeRe: showing its a cauchy sequence
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.

- October 23rd 2011, 08:51 AMDevenoRe: showing its a cauchy sequence
indeed i did! good catch. :)