# 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.

# Extending L' Hopital. (For my students. )

How can one deal with $\lim_{x \to 0^+} x^{x^2}$? It appears to be the indeterminate form $0^0$. The graph shows that it has a nice limit:

(Notice, how this function produces imaginary vales when  it is between -1 and 0. )

But, this is the wrong kind of indeterminate form for L'Hopital.  If we could somehow show that it was, in fact, the same as $\frac{0}{0}$ we could use the derivative to help us... Here is an extension of L'Hopital that will do just that:

$\lim_{x \to c} f(x)^{g(x)} = \exp \lim_{x \to c} \frac{\ln f(x)}{1/g(x)}$

See if you can apply it. We will talk more about this in class. If you simply said "we can't use L'Hopital" that is correct for this problem based on what you knew up until now. If you applied it anyway and tried to say that:  $\lim_{x \to c} f(x)^{g(x)} =\lim_{x \to c} f'(x)^{g'(x)}$ ... then that is incorrect! (And I believe it leads to the wrong answer here.)