# Prove that $a_n$ is eventually zero?

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.

• How did you solve it in the original case ? Mar 14, 2021 at 19:33
• 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. Mar 14, 2021 at 19:44

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

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

• 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. Mar 14, 2021 at 20:55
• 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 $0$.... Mar 14, 2021 at 20:57
• 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 $a_N<1$.
– NN2
Mar 14, 2021 at 21:06