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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?

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    @Golbez that doesn't diverge.2012-09-19

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Here's a related question with great answers on MO:

Nonexistence of boundary between convergent and divergent series?

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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 $G$.

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    Basically, yes. Some artificial counterexample could be produced (like $Q(x) = \sqrt{x} + \sin(e^x)$), but for all natural choices of $Q$ what you say is true.2012-09-20
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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 and note that $g\in o(f)$ yet $\sum_{i=0}^\infty g(i)\geq \sum_{j=1}^\infty \frac{1}{j}=\infty$.