Curious limit of a sequence used to prove Etemadi's SLLN I've been struggling with the proof of this proposition:

Let $\{x_{n}\}_{n=1}^{\infty}$ be a sequence of non-negative real numbers. If $$\lim_{n\to\infty} \frac{1}{\left\lceil{r^n}\right\rceil}\sum\limits_{i=1}^{\lceil{r^n}\rceil}x_{i} = c$$ for all $r >1$ with $r\in \mathbb{R}$,
  then $$\lim_{n\to\infty} \frac{1}{n}\sum\limits_{i=1}^{n}x_{i} = c.$$

I 've been researching and looking for possible fancy properties of the ceiling function that could be suitable for this proof, but all my efforts have been in vain. If you could help me out I'd be very grateful.
Observation: $\lceil{x}\rceil$ is the smallest integer $m$ such that $x \leq m$, where $x\in\Bbb R$.
Further context(taken from comment): I think this lemma/proposition is used to prove Strong Law of Large Numbers of Etemadi (1981); it can be checked between pages 55-57 of Durret's Probability: Theory and Examples (above all in page 57) but I can't write the complete formal proof of this lemma used there.
 A: The mentioned paper of Etemadi is this one. I found a free copy by search engine, but I'm not sure if it is legal so I will not include a link here. (I can't find the result in Durrett, but the book in question has at least 4 editions.) In addition to the already linked question, there is even this deleted question. Someone's class has a lot of MSE users?
Anyway. Fix an arbitrary $\alpha>1$. For every natural $k\ge\lceil \alpha \rceil $, there is an $n=n(k)$ such that $k$ belongs in one of the following intervals $$\lceil \alpha^n \rceil \le k \le  \lceil \alpha^{n+1} \rceil -1.$$
Thus by $x_i\ge 0$,
$$ \sum_1^{\lceil \alpha^{n} \rceil} x_i  \le \sum_1^{k} x_i \le \sum_1^{\lceil \alpha^{n+1} \rceil} x_i,  $$
and therefore,
$$ \frac1{\lceil \alpha^{n+1} \rceil-1}\sum_1^{\lceil \alpha^{n} \rceil} x_i  \le \frac1k\sum_1^{k} x_i \le \frac1{\lceil \alpha^{n} \rceil}\sum_1^{\lceil \alpha^{n+1} \rceil} x_i.  $$
Actually, we want a slightly different inequality. Note that    $\lceil \alpha^{n+1} \rceil  \le \lceil \alpha \lceil \alpha^n \rceil \rceil \le {\lceil \alpha^{n} \rceil }\alpha+1$. This means that we have the more relevant inequalities
$$\frac{1}{\alpha \lceil \alpha^{n}\rceil} \sum_1^{\lceil \alpha^{n} \rceil} x_i   \le \frac1k\sum_1^{k} x_i \le  \frac{\alpha \lceil \alpha^{n+1} \rceil}{\lceil \alpha^{n+1} \rceil-1}\frac1{\lceil \alpha^{n+1} \rceil}\sum_1^{\lceil \alpha^{n+1} \rceil} x_i .$$
As $k\to\infty$, $n(k)\to\infty$. Thus, we obtain
$$ \frac c{\alpha}=\lim_{n\to\infty} \frac1{\alpha \lceil \alpha^{n} \rceil}\sum_1^{\lceil \alpha^{n}\rceil} x_i=\lim_{k\to\infty} \frac1{\alpha \lceil \alpha^{n(k)} \rceil}\sum_1^{\lceil \alpha^{n(k)} \rceil} x_i=\liminf_{k\to\infty} \frac1{\alpha \lceil \alpha^{n(k)} \rceil}\sum_1^{\lceil \alpha^{n(k)} \rceil} x_i  \le \liminf_{k\to\infty} \frac1k\sum_1^{k} x_i .$$
In an analogous manner, using the fact that  $\lim_{n\to\infty} \frac{\lceil \alpha^{n+1}\rceil}{\lceil \alpha^{n+1}\rceil-1}=1$, we can control the limsup by $\alpha c$. Since a liminf is bounded by the corresponding limsup, we have proven that
$$ \frac c\alpha \le  \liminf_{k\to\infty} \frac1k\sum_1^{k}x_i \le \limsup_{k\to\infty} \frac1k\sum_1^{k}x_i \le \alpha c.$$
Since $\alpha>1$ is arbitrary, we conclude $\liminf_{k\to\infty} \frac1k\sum_1^{k}x_i = \limsup_{k\to\infty} \frac1k\sum_1^{k}x_i = c$, and therefore
$$ \lim_{k\to\infty} \frac1k\sum_1^{k}x_i = c.$$
