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It is a known fact that $\left\{ \sum_{i=1}^n \frac1i \right\}$ is divergent. It is also known that for any $\epsilon > 0$, $\left\{ \sum_{i=1}^n \frac{1}{i^{1+\epsilon}} \right\}$ is convergent. What about the middle ground - sequences of the form $\sum_{i=1}^n \frac{1}{i^{1 + f(i)}}$ where $f(i)$ is small. For instance, for the choice of $f(i) = i^{-1}$, the sequence, \begin{equation*} \sum_{i=1}^n \frac{1}{i^{(1 + i^{-1})}} \end{equation*} is divergent. And so is the sequence with the choice of $f(i) = 1 / \log i$. Is there any characterization of the functions that make this sequence convergent and divergent? What about if $f (i)$ is the reciprocal of the inverse Ackermann function?

Television
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    A simple result would come from comparing $i^{1+f(i)}$ to $i\mbox{log}(i)$, i.e. to compare $f(i)$ to $\frac{\mbox{log}\big(\mbox{log}(i)\big)}{\mbox{log}(i)}$. The reciprocal of the inverse Ackerman function goes to infinity slower than $\frac{\mbox{log}(i)}{\mbox{log}\mbox{log}(i)}$, so with $f$ the reciprocal of inverse Ackerman function for instance, it diverges – charmd Feb 20 '19 at 11:47
  • @CharlesMadeline Ah. I was not aware of this behaviour of $i \log i$. How can one prove this result? – Television Feb 20 '19 at 11:56
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    Compare to an integral: because the function $\frac{1}{x \mbox{log}(x)}$ is decreasing, $\frac{1}{i\mbox{log}(i)} \ge \displaystyle{\int_i^{i+1}} \frac{dx}{x\mbox{log}(x)}$ so $\sum \limits_{i=3}^N \frac{1}{i\mbox{log}(i)} \ge \displaystyle{\int_3^{N+1}} \frac{dx}{x\mbox{log}(x)} = \mbox{log}(\mbox{log}(N+1)) - \mbox{log}(\mbox{log}(3)) \underset{N \to \infty}{\longrightarrow} +\infty$. It is quite classic (see https://math.stackexchange.com/questions/574503/infinite-series-sum-n-2-infty-frac1n-log-n) – charmd Feb 20 '19 at 12:00
  • @CharlesMadeline By the same argument, the partial sum $\sum \frac{1}{i \log i \log \log i}$ also diverges. In which case your assertion is loose in that every function slower than $i \log i \log \log i$ also diverges. Or for that matter $i \log i \log \log i \log \log \log i \cdots$. And it is still not clear that this is the tightest characterization. But this is still a useful bit of information. Thank you. – Television Feb 20 '19 at 12:10
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    Yes, this is a good remark ! I do not think there is such a thing as a 'tightest characterization' (it doesn't really work well to find a 'slowest increasing sequence' usually). What you now know however is that the sum of reciprocals of $i\times \mbox{log}(i) \times \mbox{log log}(i)\times\cdot\cdot\cdot\times\mbox{log...log}(i)^{\alpha}$ converges or diverges depending on whether $\alpha>1$ or not – charmd Feb 20 '19 at 16:23
  • Yes. That is a good result to expect. Thanks! – Television Feb 21 '19 at 14:02

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