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What can be said about the convergence of the following modification of the hyperharmonic series ($\sum_{n=1}^{\infty} \frac{1}{n^{s}}$, which is convergent for any s>1): $$\sum \frac{1}{n^{s_n}}$$ with $s_n$ strictly monotonically approaching 1 from above? In case both convergence and divergence are still possible under this condition, is it possible to give a specific criteria for convergence, e.g. in terms of the rate of convergence of $s_n$?

Raphael J.F. Berger
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    It certainly might not converge. For example $s_n=\log_n(n+1)$. – Thomas Andrews Mar 28 '16 at 18:46
  • So the question is, can you make $s_n$ converge slowly enough to $1$ to make it converge. – Thomas Andrews Mar 28 '16 at 18:52
  • Yes, this is the question which is left over, after that on-spot counterexample! – Raphael J.F. Berger Mar 28 '16 at 18:52
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    I mean, there are cases of $s_n$ where the series doesn't converge. Probably "certainly" and "might" shouldn't be in the same sentence, ever. :) – Thomas Andrews Mar 28 '16 at 18:54
  • $s_n = \frac{log(n) + 1}{log(n)}$ would be quite slow already. Not sure if this converges. – Raphael J.F. Berger Mar 28 '16 at 19:43
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    Nope, $\frac{\log(n)+1}{\log(n)} = 1+\log_n(e)$ so $n^{s_n}=en$, and $\sum\frac{1}{en}$ does not converge. – Thomas Andrews Mar 28 '16 at 20:10
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    Nope! $$1/n^{\frac{\log n +1}{\log n}} = 1/en$$ whose sum diverges. You could use $$s_n = \frac{\log{(n\log^2 n)}}{\log n}$$ This gives us $$\sum \frac{1}{n\log^2 n} $$ which converges – Kitegi Mar 28 '16 at 20:10
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    So thats great! So all in all it means it really does depend on how fast $s_n$ converges to 1. – Raphael J.F. Berger Mar 28 '16 at 20:14
  • Would it be possible to quantify the required convergence rate exactly for obtaining a sharp criteria? – Raphael J.F. Berger Mar 28 '16 at 20:31
  • I see that for $s_n = \frac{log(n;log(n))}{log; n}$ it diverges (http://math.stackexchange.com/questions/574503/infinite-series-sum-n-2-infty-frac1n-log-n). But we can put some $s_n = \frac{log(n; f(n))}{log; n}$ here, say $n^\epsilon$. – Raphael J.F. Berger Mar 28 '16 at 21:11
  • Im mean we can put there some $f(n)$ in. And this choice gives $s_n = 1 + \epsilon$, when I'm not mistaken, thus is not monotonically converging to 1. – Raphael J.F. Berger Mar 28 '16 at 21:18
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    There are no sharp boundaries in convergence. You just stare at some $f(n)$ (here or elsewhere) and wonder how fast it might grow, then you check $f(n)=n^s$ and find (say) that $s\le1$, then you throw in some slower-growing function, so now $f(n)=n\cdot\log^r n$, then you find the critical value for $r$, then it turns out that you may add some of $\log\log n$, and so it goes. – Ivan Neretin Mar 29 '16 at 07:44
  • Ivan Neretin: Can this process then not be understood as a kind of nested intervals - which would give rise to the existence of some unique entity? – Raphael J.F. Berger Mar 29 '16 at 14:28

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Depending on the specific choice of the sequence $s_n$, for $n\rightarrow\infty$ both convergence, i.e. for $$s_n = \frac{log(n\;log^2 n)}{log\;n} \rightarrow 1^+$$ resulting in $$\sum \frac{1}{n\;log^2 n} \rightarrow c \lt \infty;$$ or divengence, i.e. for $$s_n = log_n(n+1) \rightarrow 1^+$$ resulting in $$\sum \frac{1}{n+1} \rightarrow \infty $$ is possible.

Raphael J.F. Berger
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