Supose $\{ a_{n} \}_{n=1}^{\infty} \in \mathbb{R}^+$ and also $\lceil x \rceil$ is the ceil function. I want to prove the following.
If $\forall r > 1$ $\lim_{n \to \infty} \frac{1}{\lceil r^n \rceil} \sum_{i = 1}^{\lceil r^n \rceil} a_{i} = a$ $\implies$ $\lim_{n \to \infty} \frac{1}{n} \sum_{i = 1}^{n} a_{i} = a.$
I am looking for a formal Epsilon-based demo, but I am quite stuck with this. Here is what I tried, I have just started with the formal limit definition. But I don't know what to do with the ceil function.
Knowing $\forall r > 1$ then, $\forall \epsilon > 0, \exists N s.t.\forall n \ge N \implies \left| \frac{1}{\lceil r^n \rceil} \sum_{i = 1}^{\lceil r^n \rceil} a_{i} - a \right| < \epsilon $. This is my hypotesis.
I want to find an $\epsilon'$ (I supose it will be related with the first $\epsilon$). That $\forall \epsilon' > 0, \exists N' s.t.\forall n \ge N' \implies \left| \frac{1}{n} \sum_{i = 1}^{n} a_{i} - a \right| < \epsilon' $.
So changing the $\lceil r^n \rceil$ for $n$ make the progressions converge to the same limit.
