Let $\{a_n\}$ is a bounded sequence of non-negative real numbers, with $\alpha \in (0,1)$, then if $$|\{n \in \mathbb{N}: a_n \ge \frac{1}{2^n}\}| \leq 2^{n\alpha} $$
Prove that the series $\sum_{n=1}^{\infty}a_n$ converges.
My idea is to use the ceiling function to estimate the unbounded terms as follows:
$$|\{n \in \mathbb{N}: a_n \ge \frac{1}{2^n}\}| \leq 2^{n\alpha} < 2^{\lceil n\alpha\rceil}$$
Then
$$\sum_{k=1}^{\infty}a_k \le \sum_{k=1}^{2^{\lceil n\alpha\rceil}}a_k + \sum_{k=2^{\lceil n\alpha\rceil + 1}}^{\infty}a_k \leq S + \sum_{k=2^{\lceil n\alpha\rceil + 1}}^{\infty}\frac{1}{2^k} = S + \frac{2^{-2^{\lceil n\alpha\rceil + 1}}}{1 - \frac{1}{2}} < \infty$$
Where S is the partial sum of the series up to the ceiling.
Hence, the series converges.
Is my proof correct? It looks weird and did it right now at almost 1 am, so I'm not sure. Any other approaches will be interesting. Thanks.
I agree that Shalop's formulation of the problem is the correct one. You should have $|\{k\in\mathbb N: a_k\geq 2^{-n}\}|\leq 2^{\alpha n}$ for every $n$. Let $A$ be the bound of the sequence $\{a_k\}$. Then, since all the $a_k$'s are non-negative, arrange the sum in the following way: $$ \sum\limits_{k=1}^{\infty}a_k=\sum\limits_{a_k\geq2^{-1}}a_k~+~\sum\limits_{n=2}^{\infty}\sum\limits_{2^{-n}\leq a_k<2^{-n+1}}a_k~\leq 2A +\sum\limits_{n=2}^{\infty}\sum\limits_{2^{-n}\leq a_k<2^{-n+1}}2^{-n+1}\leq2A+\sum\limits_{n=2}^{\infty}2^{-n+1}2^{n\alpha}, $$ and this latter sum converges since $\alpha\in(0,1)$.