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If for a sequence $a:\mathbb{N}\rightarrow\mathbb{R} $ $$\sum_{n=0}^{\infty}a_n\in\mathbb{R} $$ what do we know about the convergence of $a$? It is well known that $a$ converges to $0$, but for example how fast does it converge? In general, I presume, we do not know much since with the Alternating Series Test you can construct sequences who go at any speed to $0$, but the alternating series still converges.

However, if the convergence is absolute (EDIT: it has been pointed out that this is not enough, the sequence has to be decreasing in magnitude as well), the sequence has to go to $0$ faster than $1/n$ and I suspect it either is asymptotically equivalent to $n^{\alpha}$ for some $\alpha<-1$ or it goes to $0$ much faster than any such sequence. Does anyone know if my intuition is correct or false?

Is there literature on this, some cool results? Any suggestions and enlightenments on this are much appreciated.

Manatee Pink
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    See https://math.stackexchange.com/a/452074/42969. For every converging series you can find another one whose terms are much larger and it is still converging. – Martin R Jan 29 '22 at 15:04
  • Wow, that's neat. So $na_n\to 0$ is the best we can say in the absolute convergence case, right? – Manatee Pink Jan 29 '22 at 15:22
  • @ManateePink What do you mean by that comment? Please be more precise – Adam Rubinson Jan 29 '22 at 15:27
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    @ManateePink: $n a_n \to 0$ holds for a converging series with decreasing elements but not in general. – Martin R Jan 29 '22 at 16:08
  • @Adam Rubinson, up until now I was pretty sure I proved to myself that if the partial sums converge absolutely then the sequence goes to $0$ faster than $1/n$, meaning $na_n\to 0$ as $n\to\infty$ as I mentioned in the question. However, with the recent comments I am not so sure anymore. However, with Martin R.'s comment, it seemed to me that essentially $1/n$ was the only sequence we could make conclusions about with regards to the asymptotic behavior of $a$. So, that's why I wrote the best we could do. – Manatee Pink Jan 29 '22 at 16:35
  • @Martin R., but I thought since absolutely convergent series are unconditionally convergent, we could always reorder the sequence to a decreasing one? – Manatee Pink Jan 29 '22 at 16:37
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    You talk about $na_n$. When you reorder $a_n$ (call it $b_n$), the values of $na_n$ are not the same as the values of $nb_n$. – GEdgar Jan 29 '22 at 16:58

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Your intuition is incorrect in the case of absolute convergence, you can consider logarithmic corrections to $1/n$ to make the series converge or diverge. So there is no $\alpha<-1$, and moreover there is no specific cutoff point for the growth rate either (rather, an infinite sequence of functions that straddle divergence and convergence and which become "closer" to one another in some sense, although of course always having a ratio which diverges).

pre-kidney
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    For example, $\sum 1/(n (\log n)^2)$ converges. – GEdgar Jan 29 '22 at 15:13
  • Ah, I see, yeah, I didn't think about logarithmic multipliers. Thanks for the concrete example @GEdgar. So $na_n\to 0$ is the best we can say? – Manatee Pink Jan 29 '22 at 15:20
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    For your next assignment: Find an example with $a_n > 0$, where $\sum a_n$ converges, but $n a_n$ does not go to zero. – GEdgar Jan 29 '22 at 15:22
  • @GEdgar, that comment concerns me, I was pretty sure I recently proved that result recently to myself, but looks like I have to post another question to see where I went wrong... – Manatee Pink Jan 29 '22 at 16:28
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    Hint: $a_n$ does not need to be monotonically decreasing. It could have rare terms that are relatively large. – GEdgar Jan 29 '22 at 16:57
  • You mean like, $a_n=1/n^2$, but every square $n$ $a_n=1/n$? – Manatee Pink Jan 29 '22 at 17:20