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. 

*

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


*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 $g(n) \gg f(n)$.
