Is there a "greatest function" that converges? 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. 

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 $M_k \le n < M_{k+1}$.
