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Does there exist a sequence $\{a_n\}_{n\ge1}$ with $a_n < a_{n+1}+a_{n^2}$ such that $\sum_{n=1}^{\infty}a_n$ converges?

Does there exist a sequence with the same property but with each term positive?

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    Do you have any example sequences that even converge to 0?2012-11-27
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    @user9352 Of course. When you do not impose $a_n \ge 0$, then nearly any $a_n$ works. For example, $a_n=-\frac{1}{(n+1)^2}$.2012-11-27
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    You can use $a_n = -\cfrac{1}{2^n}$ if you don't impose positiveness. Or the opposite of anything going to $0$ _fast enough_.2012-11-30
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    @xavierm02 But for $n=1, \ a_n \not < a_{n+1}+a_{n^2}$. Do you Know an example s.t. $a_n < a_{n+1}+a_{n^2}, \ \forall n \in \mathbb N?$2012-12-07
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    @Pambos: $a_n=-\cfrac{1}{2^{n+2}}$ If the first terms are the only ones bothering you, you can just truncate the sequence...2012-12-07
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    @xavierm02: I am afraid you can't. For $n=1$ the relation $a_n0$. The truncated sequence $a_n=-\frac{1}{2^{n+k}}$ doesn't satisfy this for any $k\geq2.$2012-12-08
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    @Pambos: Start with $u_n = -\cfrac{1}{2^n}$. For $n\ge2$, the inequality works. For $n=1$ to work you need $a_2>0$ so you can just pick $a_2 = 1$ then you need to make sure it works for $n=0$, ie $a_0 0$ so you can pick $a_1=1$ and then for $a_0$, you can pick anything since it doesn't influence any inequality.2012-12-08
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    @xavierm02: Then you have problem at $a_2$. You want $a_2$a_2=1$ and $a_n=-\frac{1}{2^n}, \ n\geq 3$. I believe the problem is more difficult than it seems to be. – 2012-12-08
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    @Pambos: Hm... Apparently, we can't have only negative numbers after a given index. Because we have $02012-12-08

2 Answers 2

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Okay, it is indeed impossible. First we find an $n>1$ such that $a_n$ is strictly positive. The basic idea of the proof is noting that if say $a_n

Define two functions on the positive integers, $S: m\mapsto m^2$ and $A: m\mapsto m+1$ ($S$ stands for squaring and $A$ for adding). Then recursively define the following sets: $I_0=\{n\}$ and $I_{i+1}=S(I_i)\cup A(I_i)$, i.e. $I_k$ is a $k$-fold composition of $A$s and $S$s applied to $n$. Then it follows immediately from $n>1$ that the smallest element of $I_k$ is $n+k$. I now claim that $I_k$ has $2^k$ elements. To prove it, we use induction. It's certainly true for $I_0$. If it's true for $I_{k}$, then by injectivity of $S$ and $A$, it must be the case that $S(I_{k})$ and $A(I_{k})$ both have $2^{k}$ elements. Suppose they have an element in common. Then we can write that element as a perfect square of a number that's at least $n+k$ and also either as $n+k+1$ (which is impossible - $(n+k)^2>n+k+1$ since $n+k\geq2$ by assumption) or as $u^2+i$ for some integers $u$ and $i$ with $ik+1$, which is a contradiction. Thus $S(I_{k})$ and $A(I_{k})$ are disjoint and $I_{k+1}$ has $2^{k+1}$ elements and our inductive proof is complete.

We now obtain the inequalities $a_n=\sum_{i\in I_0}a_i<\sum_{i\in I_1}a_i<\sum_{i\in I_2}a_i<....$, the only problem is that $I_i$ and $I_j$ may not be disjoint. So define a new sequence of finite index sets $J_k$ by $J_0=I_{0}$ and if $J_k$ has largest element $r$, then we define $J_{k+1}=I_r$ whose smallest element is $n+r$ so that $s\sum_{k=0}^{\infty}a_n$, so the sum diverges.

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    Nice proof! Maybe the last paragraph could be simplified a little: if $\sum_{i=1}^\infty a_i$ converges, then as $k\to\infty$, $\sum_{i\in I_k}a_i\le\sum_{i=k}^\infty a_i\to 0$, a contradiction.2012-11-27
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Suppose such a positive sequence $\{a_n\}_{n \in \mathbb{N}}$ exists.

Then by repeated application of the property $a_n < a_{n+1}+a_{n^2}$ on each term appearing starting from $a_2$, gives: $a_2 < a_3 + a_4 < a_4 + a_5 + a_9 +a_{16} < \ldots $ and so on.

Let, $\{S_k\}_{k\ge 1}$ be the resulting sequence of subscripts formed by such repeated applications. So, $S_1=\{2\},S_2=\{3,4\},S_3=\{4,5,9,16\},\ldots$ and so on.

The idea is to show that these $S_k$'s have no repeating elements, so that $\sum\limits_{n=1}^{\infty} a_n$ must diverge, because any tail of the series must be greater than $a_2 > 0$.

Now, Choose the smallest $l$, where a repetition happens in $S_l$. Then the repeating element must be of the form $m^2$, for some $m \in \mathbb{N}$, which comes from $m \in S_{l-1}$ and $m^2-1 \in S_{l-1}$.

But, then $m^2-1 \in S_{l-1} \implies m^2-2 \in S_{l-2} \implies \ldots \implies m^2-2m+2 \in S_{l-2m +2}$ since, none of $m^2-1,m^2-2,\ldots,m^2-2m+2$ are perfect squares. Thus we must have $l > 2m-2$.

Also, $m \in S_{l-1} \implies m > l-1$.

Which is absurd as $m$ is greater than $2$.

(Q.E.D.)