Let $(a_n)$ be a nonnegative sequence such that $\sum_{n=1}^{\infty}a_n < \infty$. Suppose we have the following

$a_{n+1}^2-a_n^2+\alpha a_n \le 0 \quad \forall n\in \mathbb{N}$ for some $\alpha >0$ $\quad (*)$

I can prove by contradiction that under the above assumptions $a_n$ is eventually zero, which means there exists $N \in \mathbb{N}$ such that $a_n=0 \quad \forall n \geq N$.

However I'm pretty curious about whether the result still holds when we substitute $\alpha a_n$ in $(*)$ by $\alpha a_{n+1}$ (all other assumptions remained). I've been working on this but haven't had an answer so far.

Any idea/answer would be appreciated.

  • $\begingroup$ How did you solve it in the original case ? $\endgroup$ Mar 14, 2021 at 19:33
  • $\begingroup$ As I said before, I assume that we can extract a positive subsequence $(a_{k_{l}})$ of $a_n$. Then take the sum on both sides of $(*)$ after dividing $(*)$ by $a_{k_{l}}$. Finally, we see a contradiction. $\endgroup$
    – No name
    Mar 14, 2021 at 19:44

1 Answer 1


$a_{n+1}(a_{n+1} +\alpha) \le a_n^2$

$(a_{n+1} + \alpha) \le \frac {a_n^2}{a_{n+1}}$ for the $a_{n+1} > 0$.

If we say $a_{n+1} \ge \beta a_n^2$ then $\frac {a_n^2}{a_{n+1}} \ge \frac 1\beta$ and this is certainly possible.

Let $a_{n+1} = \frac 12 a_n^2; a_1 < 1$ then we want

$\frac 12a_n^4 + \alpha \frac 12a_n^2 \le a_n^2$ which can be true for $\alpha = 1$

  • $\begingroup$ I did mean $a_k > 0$. Obviously $a_n \to 0$ (other wise the series of non-negative terms can converge) so, yes, that was a complete typo. $\endgroup$
    – fleablood
    Mar 14, 2021 at 20:55
  • 1
    $\begingroup$ Also the only reason I made a comment about anything was to divide by $a_{n+1}$. Need $a_{n+1} \ne 0$ for that. (Of course we can have $a_k$ occasionally $0$.... $\endgroup$
    – fleablood
    Mar 14, 2021 at 20:57
  • $\begingroup$ I think it suffices to take $a_{n+1} = \frac{1}{2\alpha}a_n^2$. Because $a_n \to 0$ (as $\sum a_n <+\infty$) then the condition $a_{n+1}^2-a_n^2+\alpha a_{n+1} \le 0$ always hold true. Also because $\sum a_n <+\infty$ then there exists an $N$ such that $a_N<1$. $\endgroup$
    – NN2
    Mar 14, 2021 at 21:06

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