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!

  • Are $a_k$s positive? – Mostafa Ayaz Aug 24 at 12:37
  • $(a_k)_k\ge0$, but of course you can take $(a_k)>0$ when you want to give counter example. – Tony Ma Aug 24 at 12:40
  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 $0$.

  • What is the derivative of $\log$*? (My question 2) – Tony Ma Aug 24 at 12:41
  • @TonyMa $\log^*$ isn't even continuous, but mostly has derivative zero. You can smooth it out if you wish. But the properties of this new function won't exactly be unique. – Robert Wolfe Aug 24 at 12:44
  • Yes, I know it is not continuous, I am just asking about how to solve my question 2 (do we need another function?) – Tony Ma Aug 24 at 12:46
  • @TonyMa As mentioned by Robert this function is not continuous, hence not really differentiable. Hoever the number of discontinuities are countable, so depending on the amount of mathematics you had: This funciton is almost everywhere differentiable, which is what is actually needed most of the time. The solution to the second question has been added. – Jürg Merlin Spaak Aug 24 at 12:50
  • You are right... the amount of mathematics I have is not enough... but your answer seems to be good (although I can't fully understand). – Tony Ma Aug 24 at 12:59

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