# 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.