Is there a "greatest function" that converges?

Question

We just hit convergence tests in calculus, and learned that $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges for all $p \gt 1$. I thought that this was sort of a "barrier" between what converges and what diverges. Specifically, that setting $a_n=\frac{1}{n^{1+\epsilon}}$ is sort of the "greatest function" (I'll make this precise later) for which $\sum a_n$ converges.

But, I did realize that there are functions that dominate $\frac{1}{n^{1+\epsilon}}$ but not $\frac1n$, such as $\frac{1}{n\log(n)}$. Now, the sum of that specific example diverges, but it got me wondering about whether $\frac{1}{n}$ is truly the "boundary". So, this leads me to two questions.

1) Is there a function $f$ that dominates $\frac{1}{n^p}$ for all $p>1$, meaning: $$\lim_{x\to\infty} \frac{f(x)}{\frac{1}{x^p}}=\infty$$ Such that: $$\sum_{n=1}^\infty f(n)$$ converges?

2) If so, up to a constant is there a function $g$ such that $\sum_{n=1}^\infty g(n)$ converges, such that $g$ dominates $f$ for all other functions $f$ such that $\sum_{n=1}^\infty f(n)$ converges?

I'm just a freshman in high school so I apologize if this is a stupid question.

Answer 1

1. Yes, $f(n)=\frac{1}{n(\ln{n})^2}$.

2. No. Assume $f \geq 0$ and $\sum_{n \geq 1}{f(n)} < \infty$.

Then there exists an increasing sequence $N_n$ and some constant $C > 0$ such that $\sum_{N_n+1}^{N_{n+1}}{f(k)} \leq C2^{-n}$.

Then set $g(n)=(p+1)f(n)$, where $N_p < n \leq N_{p+1}$.

Then $\sum_{N_n+1}^{N_{n+1}}{g(k)} \leq C(n+1)2^{-n}$, thus $\sum_{n \geq 1}{g(n)}$ is finite and $g(n) \gg f(n)$.

Answer 2

1) $$f(n) = \frac{1}{n \log(n)^2}$$

2) No. Given any $g > 0$ such that $\sum_n g(n)$ converges, there is an increasing sequence $M_k$ such that $$\sum_{n \ge M_k} g(n) < 2^{-k}$$
Then $ \sum_n g(n) h(n)$ converges, where $h(n) = k$ for $M_k \le n < M_{k+1}$.