Re: Show a sequence does not have a limit

On Mon, 20 Oct 2008, Cliff MacGillivray wrote:

Show x_n=1 + 1/sqrt(2) + ... + 1/sqrt(n) does not have a limit using the
Cauchy Criterion.

Why do it the hard way? Compare it with sum(n=1,oo) 1/n.

Solution:
If n>m then we have s_n - s_m =
=1/sqrt(m+1) + 1/sqrt(m+2) + ... + 1/sqrt(n)
> 1/sqrt(n) + 1/sqrt(n) + ... + 1/sqrt(n)
= (sqrt(n) - sqrt(m))/sqrt(n)
since we have n-m 1/sqrt(n) terms
=1 - (sqrt(m) / sqrt(n))

So, if we choose n=4m than we have s_4m - s_m > 1/2 . Thus the sequence
cannot be cauchy since we cannot choose an aribitary h >0 such that for
any n,m |s_n - s_m| < h and so it does not have a limit.

Correct?

Yes.

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