Almost Convergence.

The title of my last post "Almost Cauchy Sequence" made me wonder if "Almost Cauchy" was already defined in some formal sense. I could not find much on "Almost Cauchy" ... but I did find out about "Almost Convergence." (I think I have ran in to this before... but it is very neat... observe:)

A sequence S = S_n is almost convergent to L if for any \epsilon > 0 we can find an integer n such that the average of n or more consecutive terms in the sequence is within \epsilon of L. Formally, \forall \epsilon > 0 \quad \exists N \ni \quad \displaystyle \left| \frac{1}{n} \sum_{i=0}^{n-1} S_{k-i} -L    \right| < \epsilon \quad \forall n > N, \; k \in \mathbb{N}. (I took this definition from here, and I wonder why they didn't mention that we need k \geq n or else we are looking at negative terms in the sequence... But, maybe there is an interpretation of something like S_{-3}? Or maybe the author thinks that would be obvious? I'm going to ask at the mathematics stack exchange!)

Almost Cauchy Sequence, unfinished business.

I just remembered a question from beginning analysis that my professor mentioned in class:

"Find a sequence A that is not Cauchy, yet any consecutive terms A_n, A_{n+1} will become arbitrarily close."

It was one of those questions that he mentioned in class for us to do on our own, but I never quite got to it. I was thinking about convergence and complete metric spaces today (and about unfinished business, as I've made a lot of my own since I started teaching.) And the question came back to me.

We seek a sequence, A, where \forall \; \epsilon > 0 \quad \exists N \in \mathbb{N} \;  \ni \; \forall n > N \quad |A_n - A_{n+1}| < \epsilon and \exists \epsilon > 0 \; \ni  \quad \forall  N \in \mathbb{N} \quad \exists m,n > N \quad |A_m - A_n| > \epsilon. (Writing the negation of what it means for a sequence to be Cauchy was a little more tricky than I thought...)

The natural log function is the most obvious answer. For large vales of n \ln (n) - \ln (n+1) is small. That is, \displaystyle \lim_{n \rightarrow \infty} \ln \frac{n}{n+1} = 0. This is true becuase after taking EXP of both sides we have: \displaystyle \lim_{n \rightarrow \infty}  \frac{n}{n+1} = 1

Next, it seems obvious that the natural log can't be Cauchy, but just to be certian if we let \epsilon = \frac{1}{2} then, \forall N \in \mathbb{N} if we choose some n > N and let m= ne (so, n,m > N) then

| \ln m - \ln n | = |\ln \frac{m}{n}| = |\ln \frac{ne}{n}| = 1 > \frac{1}{2} = \epsilon.

NOT Cauchy at all!

Ok, I imagine my professor was thinking of the natural log... but, what other functions also have this property? \sqrt[n]{x}? So, I guess this is really asking when \displaystyle \lim_{x \rightarrow \infty} f(x) - f(x+1) = 0 while \displaystyle \lim_{x \rightarrow \infty} f(x) = \pm \infty. It is necesairy but not sufficent for f'(x) > 0 \; \forall x and f''(x) < 0 \; \forall x , both will have the x-axis as a horizontal asymptote.

Nice to have finally gotten around to doing this problem! I think it will be good practice to find more unfinished business in my notes.