Is there a minimal diverging series? Is there a function $f:\mathbb{N} \to \mathbb{R}^+$ s.t. its series $\Sigma_{i=0}^\infty f(n)$ diverges but the series for all function in $o(f)$ converge?
 A: Sadly, no such function can exist. For suppose $f$ is such a function. Define the partial sums:
 $$F(n) := \sum_{i=0}^n f(n) $$
Then you can find a function that diverges more slowly, say:
 $$ G(n) := \sqrt{F(n)} $$
This new $G$ is increasing, so you it is the sequence of the partial sums for $g$ given as:
$$ g(n) = G(n) - G(n-1) $$
It remains to check that $g = o(f)$:
$$ \frac{g(n)}{f(n)} = \frac{G(n) - G(n-1)}{G(n)^2 - G(n-1)^2} = \frac{1}{G(n) + G(n-1)} < \frac{1}{G(n)} = o\left(1\right)$$
where the $o(1)$ approximation follows from divergence of $G$.
A: No. Suppose $\sum_{i=0}^\infty f(i)=\infty$. Then we can find values $N_j\in \mathbb N$ such that 
$$\sum_{i=N_j}^{N_{j+1}-1}f(i)>1$$
for all $j\in\mathbb N$. Define $g:\mathbb N\to\mathbb R^+$ by
$$g(i)=\begin{cases}f(i) &\text{if } i<N_1\\
\frac{1}{2}f(i) &\text{if } N_1\leq i<N_{2}\\
\frac{1}{3}f(i) &\text{if } N_2\leq i<N_{3}\\
&\vdots
\end{cases}$$
and note that $g\in o(f)$ yet $\sum_{i=0}^\infty g(i)\geq \sum_{j=1}^\infty \frac{1}{j}=\infty$.
A: Here's a related question with great answers on MO:
Nonexistence of boundary between convergent and divergent series?
