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:

$$|\{n \in \mathbb{N}: a_n \ge \frac{1}{2^n}\}| \leq 2^{n\alpha} < 2^{\lceil n\alpha\rceil}$$


$$\sum_{k=1}^{\infty}a_k \le \sum_{k=1}^{2^{\lceil n\alpha\rceil}}a_k + \sum_{k=2^{\lceil n\alpha\rceil + 1}}^{\infty}a_k \leq S + \sum_{k=2^{\lceil n\alpha\rceil + 1}}^{\infty}\frac{1}{2^k} = S + \frac{2^{-2^{\lceil n\alpha\rceil + 1}}}{1 - \frac{1}{2}} < \infty$$

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.


closed as unclear what you're asking by Nick Peterson, Shalop, user91500, Gabriel Romon, user223391 Sep 8 '17 at 18:07

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  • $\begingroup$ I'm just wondering if I can say that the rest of the terms are less than $\frac{1}{2^n}$ $\endgroup$ – Richard Clare Sep 4 '17 at 4:18
  • $\begingroup$ I'm not quite sure I follow. In $\lvert\{n\in\mathbb{N}: a_n\geq\frac{1}{2^n}\}\rvert\leq 2^{n\alpha}$, where does the $n$ outside of the magnitude come from? How is it defined? $\endgroup$ – Nick Peterson Sep 4 '17 at 4:20
  • $\begingroup$ @NickPeterson I guess is the same of the n-th term... Is from various past qualifying exams It does not say anything else. $\endgroup$ – Richard Clare Sep 4 '17 at 4:21
  • $\begingroup$ The $n$ inside the set notation has no meaning outside of it. As stated, this is meaningless. And without understanding the condition given, it is awfully hard to use it to prove anything. $\endgroup$ – Nick Peterson Sep 4 '17 at 4:22
  • 3
    $\begingroup$ @NickPeterson I suspect that the correct formulation of the problem should be that $\# \{ k \in \Bbb N: a_k \geq 2^{-n} \} \leq 2^{\alpha n}$, for every $n$. In this case the claim is true. Then again, I can't say for certain what it should or should not say. As currently stated, you are right that the problem does not make any sense (hence neither does the solution). $\endgroup$ – Shalop Sep 4 '17 at 4:50

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 $\alpha\in(0,1)$.

  • $\begingroup$ Thank you. Very short and simple solution. $\endgroup$ – Richard Clare Sep 4 '17 at 23:34

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