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Does there exist a sequence $\left(a_n\right)_{n\ge1}$ with $a_n < a_{n+1}+a_{n^2}, \forall n=1,2,3,\ldots$ such that the series $\displaystyle{\sum_{n=1}^{\infty}a_n}$ converges?

This is the first part of this question which has an (accepted) answer for its second part only:
The last sentence of the answer is:
"Now we note that $\sum_{i=1}^{\infty}a_i\geq\sum_{k=0}^{\infty}\sum_{i\in J_k}a_i>\sum_{k=0}^{\infty}a_n$, so the sum diverges."
For the inequality $\sum_{i=1}^{\infty}a_i\geq\sum_{k=0}^{\infty}\sum_{i\in J_k}a_i$ to be valid, we have to assume the positivity of $(a_n)_{n\in\mathbb N}$ since $\displaystyle{\bigcup_{k\in\mathbb N}J_k\neq\mathbb N}$.

According to the comments the first part is a difficult question.

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@mjqxxxx: For $n=1$ the inequality $a_n<a_{n+1}+a_{n^2}$ becomes $a_2>0$ so the sequence cannot be always negative. –  George Jan 3 '13 at 23:50
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This is one of those graveyard questions -- four deleted answers within one hour. –  joriki Jan 4 '13 at 0:01
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@user7530: The sequence $(a_n)_{n\in\mathbb N}$ must contain a strictly positive subsequence. Can you explain more how non-existence follows from the linked proof? I believe it doesn't. –  P.. Jan 4 '13 at 0:07
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@RossMillikan: This question has an accepted answer only for its second part. Not for the one I am asking. If this is against the rules of the site I will delete it. –  George Jan 4 '13 at 0:13
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@user7530: since this question was closed by the community, the standard thing to do to request for re-open votes is to open a re-open request like this on meta. –  Willie Wong Jan 4 '13 at 8:33
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1 Answer

up vote 13 down vote accepted
+100

Yes, if negative elements are allowed, the series can be made to converge. The basic idea is to first fix the values at the square positions, and then use the remaining freedom to create subsequences between those positions that sum to $0$ and whose partial sums are bounded by a decreasing function. Specifically, for each $k \ge 1$ and $0 \le l \le 2k$, let $a_{k^2 + l}=b_{k}(1- c_{k,l})$, where $(b_k)_{k\ge 1}$ is a positive sequence that converges monotonically to zero, $c_{k,0}=0$ (so $a_{k^2}=b_{k}$), and $c_{k,l}$ is non-decreasing for each $k$. Then for each $k \ge 1$ and $0 \le l < 2k$ we have the "internal constraints," $$ b_{k}(1-c_{k,l})=a_{k^2+l}<a_{k^2+l+1}+a_{(k^2+l)^2}=a_{k^2+l+1}+b_{k^2+l}=b_{k}(1-c_{k,l+1})+b_{k^2+l}, $$ and for each $k \ge 1$ we have the "join constraint," $$ b_{k}(1-c_{k,2k}) = a_{k^2+2k}<a_{k^2+2k+1}+a_{(k^2+2k)^2}=b_{k+1}+b_{k^2+2k}. $$ We will guarantee the internal constraints are met by imposing a stricter constraint for each $k$: $$ \max_{0\le l < 2k}(c_{k,l+1}-c_{k,l})\le \frac{b_{(k+1)^2}}{b_{k}}; $$ using this, we see that $$ c_{k,l+1}-c_{k,l}\le \max_{0\le l < 2k}(c_{k,l+1}-c_{k,l})\le \frac{b_{(k+1)^2}}{b_{k}} < \frac{b_{k^2+l}}{b_{k}} $$ does hold for each $k$ and $l$. The join constraint is met if $$c_{k,2k} > 1-\frac{b_{k+1}+b_{k^2+2k}}{b_k};$$ in particular, since $(b_{k})$ is positive, it's sufficient to have $c_{k,2k}\ge 1$. Now, for $k>1$ we can take $c_{k,l}=1$ except for $c_{k,0}=0$, $c_{k,1}=1/2$, $c_{k,2k-1}=3/2$, and $c_{k,2k}=2$. For $k=1$ we will take $c_{k,0}=0$, $c_{k,1}=1/2$, and $c_{k,2}=1$. Then the join constraint is met, and the internal constraint becomes $$ \frac{b_{(k+1)^2}}{b_{k}} \ge \max_{0\le l < 2k}(c_{k,l+1}-c_{k,l})=\frac{1}{2}, $$ which limits how quickly $(b_{k})$ can converge to $0$. For example, we can choose $$ b_{k}=\frac{1}{\sqrt{\log(k+1)}}. $$ The resulting sequence does what we want. For each $k>1$, the subsequence from position $k^2$ to position $(k+1)^2-1$, inclusive, sums to zero, and its partial sums are never larger than $\frac{3}{2}b_{k}$, a bound that decreases (albeit slowly) to zero as $k\rightarrow\infty$. The overall series, therefore, converges to the sum of the first three terms, which is $\frac{3}{2}b_1$. The sequence consists of a series of EEG-like "blips" of four non-zero values at the square positions; each blip has a slightly smaller amplitude than the last.

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+1, very nice. Just an irrelevant detail: The value of $a_1$ is actually arbitrary, so $c_{k,0}=1$ is OK, but from your announcement that $c_{k,l}$ is non-decreasing for each $k$ and from the sum $\frac32b_1$ for the first three terms it seems that you intended $c_{k,0}=0$? –  joriki Jan 6 '13 at 8:41
    
This answer is a model of clarity and concision. –  Ewan Delanoy Jan 6 '13 at 8:44
    
@joriki: Thanks, that was indeed a typo. –  mjqxxxx Jan 6 '13 at 21:49
    
Nice answer! Thanks. –  George Jan 11 '13 at 20:18
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