Existence of a sequence $\{\epsilon_n\}_{n\ge 1}$ such that $\sum\limits_{n=1}^{\infty}\frac{1}{n^{\varepsilon_n}} $ converges

Question

I am trying to find a sequence $\{\varepsilon_n\}_{n\ge 1}$ such that

$$\lim_{n\to \infty}\varepsilon_n=1~~~~~~~~~~~~~\text{and}~~~~~~~~~\sum_{n=1}^{\infty}\frac{1}{n^{\varepsilon_n}} <\infty.$$

In case such sequence $\{\varepsilon_n\}_{n\ge 1}$ exists I would like to have an explicit example.

Remark: This is an interesting problem since we know that for the case where $\varepsilon_n = 1$ we have

$$\sum_{n= 1}^{\infty} \frac{1}{n^1}=\infty$$

Answer 1

Since $\sum_{n=1}^{\infty}\frac{1}{n\log^2 n} $ converges, if we choose $n^{\varepsilon_n} =n\log^2 n $ or $n^{(\varepsilon_n-1)/2} =\log n $ or $(\varepsilon_n-1)/2 =(\log \log n)/\log n $ or $\varepsilon_n =1+(2\log \log n)/\log n $, the resulting series will converge and $\lim_{n \to \infty} \varepsilon_n = 1 $.

Similarly, if $\varepsilon_n =1+(\log \log n)/\log n $, the resulting series will diverge.

Answer 2

Thoughts, but definitely not an answer:

A standard proof that $\sum \frac{1}{n}$ diverges is to group things in groups that are size a power of $2$. (1) + (1/2 + 1/3) + (1/4 + ... + 1/8) + ... and then argue that each element in a list is at least as big as the last one, and therefore each list has a sum that's at least $1/2$.

A kind of "dual" of that is to say that each element is no \emph{larger} than the first, and therefore each group is no more than $2^k$ times the first element (where the group has size $2^k$). If only those numbers ($2^k$ times the first element) were well-behaved, this would get you an UPPER sum for the series and you could show it was bounded. It's not, of course, but there's half an idea there.

Suppose your numbers $\epsilon_n$ approached $1$ from above slowly enough that each group (in some grouping like the one above) had a sum that was manageable. Then you could actually prove boundedness. One thing you realize is that the decrease in your exponents has to be more or less exponentially slow.

Answer 3

Yes, you can find such $\epsilon_n$. First, the series below is convergent $$ \sum_{n\geq 2} \frac{1}{n (\ln n)^{1+s}} $$ for any $s>0$. We choose $s=1$ and $\epsilon_n$ such that $2n(\ln n)^2>n^{\epsilon_n}>n(\ln n)^2$ for any $n\geq 2$. We can find that $\epsilon_n\to 1$ as $n\to \infty$.