# showing its a cauchy sequence

• Oct 22nd 2011, 11:46 AM
Duke
showing its a cauchy sequence
Show $\displaystyle i^n/n$ is a cauchy sequence. need $\displaystyle |(mi^n-ni^m)/nm| < e$whenever $\displaystyle n,m>_N$.
• Oct 22nd 2011, 05:56 PM
Ackbeet
Re: showing its a cauchy sequence
Silly question, I suppose, but does $\displaystyle i=\sqrt{-1}$ in this problem? If so, why not just use the triangle inequality?
• Oct 22nd 2011, 05:57 PM
Drexel28
Re: showing its a cauchy sequence
The sequence converges, and so it's trivially Cauchy. It converges to zero because each term has modulus $\displaystyle \frac{1}{n}$.
• Oct 23rd 2011, 12:01 AM
Deveno
Re: showing its a cauchy sequence
true, but the nice thing about cauchy sequences is that there's no limits involved. note that:

$\displaystyle \left|\frac{i(m- n)}{mn}\right| = |i|\left|\frac{m-n}{mn}\right| = \left|\frac{m-n}{mn}\right|$

$\displaystyle \leq \left|\frac{1}{n}\right| + \left|\frac{-1}{m}\right| = \left|\frac{1}{n}\right| + \left|\frac{1}{m}\right|$

if we choose $\displaystyle N > 2/\epsilon$, then for m,n > N

$\displaystyle \left|\frac{1}{n}\right| + \left|\frac{1}{m}\right| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

NOTE: if i represents a constant, rather than the square root of -1, then just choose:

$\displaystyle N > \frac{2|i|}{\epsilon}$
• Oct 23rd 2011, 03:23 AM
Duke
Re: 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. $\displaystyle |\frac{i^n}{n}-\frac{i^m}{m}|<_ |\frac{i^n}{n}|+|\frac{i^m}{m}|=\frac{1}{n}+\frac{ 1}{m}$ Then choosing N >2/e completes the proof.
• Oct 23rd 2011, 08:51 AM
Deveno
Re: showing its a cauchy sequence
indeed i did! good catch. :)