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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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Take $f(n)=1/2^n$ – Golbez Sep 19 '12 at 6:55
@Golbez that doesn't diverge. – Kaveh Sep 19 '12 at 6:57

3 Answers 3

up vote 6 down vote accepted

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 $\sum_{i=0}^\infty g(i)\geq \sum_{j=1}^\infty \frac{1}{j}=\infty$.

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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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It seems to me that the construction should work not only for $\sqrt{.}$ but more generally for any function in $o(n)$, right? – Kaveh Sep 19 '12 at 7:14
ps: it also needs to be unbounded. Are we using any other property of $\sqrt{.}$? – Kaveh Sep 19 '12 at 7:26
The function used in place of $\sqrt{\cdot}$ should certainly be unbounded, increasing, $o(n)$ and it should map $0$ to $0$. I did use some peculiarities of $\sqrt{\cdot}$ when I was estimating the order of growth of $\frac{g(n)}{f(n)}$, but I think the same argument will work for any well behaved function. – Jakub Konieczny Sep 19 '12 at 8:09
So we also need the function $Q=\sqrt{.}$ to be s.t. $\frac{Q(F(n))-Q(F(n-1))}{F(n)-F(n-1)} \in o(1)$. – Kaveh Sep 19 '12 at 15:37
Yes, indeed. If you assume that $Q$ is defined for all positive reals (as opposed to integers) and reasonably smooth, you can use Lagrange to prove that $$( Q(F(n)) - Q(F(n-1)) )/(F(n) - F(n-1)) = Q'(y_n)$$ with some $y \in (F(n),F(n-1))$. So what you really need is $\lim_{x \to \infty} Q'(x) = 0$, which will normally be easier to verify. – Jakub Konieczny Sep 19 '12 at 20:23

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