Function $f$ s.t. $\lim_{x\to\infty}\frac{f(e^x)}{f(x)}=1$

The questions are:

1) Does there exists some function $f$ s.t. $\lim_{x\to\infty}\frac{f(e^x)}{f(x)}=1$ and $\lim_{x\to\infty}f(x)=\infty$?

2) Is $\big(\sum_{k=n}^{2^n}a_k\big)_n\to0$ is sufficient to guarentee the convergence of $\sum_{n=1}^\infty a_n$ for $(a_n)_{n\in\mathbb > N}\ge0$?

I recently discovered that $(a_n)_{n\in\mathbb N}\ge 0$ and $\big(\sum_{k=n}^{2n}a_k\big)_{n\in\mathbb N}\to 0$ doesn't implies $\sum_{n=1}^\infty a_n\in\mathbb R$, which is done by setting $(a_n)_{n\ge2}=\frac{1}{n\ln n}$.

By setting $(a_n)=\frac{1}{n\ln n \ln\ln n}$, we can't guarentee convergence of $\sum_{n=1}^\infty a_n$ even if $\big(\sum_{k=n}^{n^2}a_k\big)_n\to 0$

But what after $2n$ and $n^2$ is $2^n$ (in some sense), so I would like to ask whether $\big(\sum_{k=n}^{2^n}a_k\big)_n\to0$ is sufficient to guarentee the convergence of $\sum_{n=1}^\infty a_n$ for $(a_n)_{n\in\mathbb N}\ge0$. Of course it is necessary, but I don't believe it is sufficience. And thus I raise question 1.

Answering question 1 help question 2 of course, but is it sufficient?

Perhaps we should consider functions (to be an indefinite integral of the integrand function-to-be) like the inverse of $g(x)=x^x$ (superlog?)? But tetration don't satisfy many properties that power have. And what is the derivative of it? Is it possible to make it simple? Or else, perhaps we should try other functions.

Any help will be appreciate. Thank you!

Solution 1:

1. Yes, there exists such a function: $\log^*$, which is defined as the minimal $n\in \mathbb{N}$ such that $\log^n (x) <1$, where $\log^n$ is the $n$-fold $\log$ (i.e. $\log(\log(\cdots))$. This is a very slowly increasing function, but eventually is infinite. Furthermore it's easy to see that we have $\lim_{x\rightarrow \infty} \frac{\log^*(e^x)}{\log^*(x)}=\lim_{x\rightarrow\infty}\frac{\log^*(x)+1}{\log^*(x)}=1$.

2. With the inverse function of $\log^*$, however we take the base $2$. We define $a_k = \frac{1}{n}\Leftrightarrow (k = \text{argmin}\{\log^*(x)=n\})$ and $0$ otherwise. We easily see, that $\sum_{k=1}^\infty a_k= \infty$. On the other hand $\sum_{k=n}^{2^n}a_k$ contains at most one nonzero $a_k$ so it tends to $0$.