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If the average of a sequence converges, can I find a uniform bound that does not depend on where I start?

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Let $\{a_k\}_{k\in \mathbb{Z}} \subset \mathbb{R}$ a real sequence and $a\in \mathbb{R}$ such that $$ \lim_{n\to +\infty} \frac{1}{n} \sum_{k=1}^n a_k = a = \lim_{n\to +\infty} \frac{1}{n+1} \sum_{k=0}^n a_{-k},$$ for all $j\in \mathbb{Z}$, the same is true for $\{a_{k+j}\}_{k\in \mathbb{Z}} \subset \mathbb{R}$.Therefore for all $\epsilon>0$ and $j\in \mathbb{Z}$ we can find a $N_{\epsilon,j}\in \mathbb{N}$ such that for all $n\geq N_{\epsilon,j}$, $$\left|\frac{1}{n} \sum_{k=1}^n a_{k+j}-a\right|< \epsilon \; \text{ and } \;\left |\frac{1}{n+1} \sum_{k=0}^n a_{-k+j}-a\right|< \epsilon.$$If we add as an additional condition that the sequence $\{a_k\}_{k\in \mathbb{Z}}$ is bounded, I would like to show that for all $\epsilon >0$ there exists $N_\epsilon \in \mathbb{N}$ such that for all $j \in \mathbb{Z}$ and $n \geq N_\epsilon$, $$\left|\frac{1}{n} \sum_{k=1}^n a_{k+j}-a\right| < \epsilon \; \text{ and } \; \left|\frac{1}{n+1} \sum_{k=0}^n a_{-k+j}-a\right|< \epsilon.$$Without this additional condition, I can show that $N_{\epsilon,j}$ is proportional to $|j|$ for $|j|$ large enough.The sequence $1,-1,\sqrt{3},-\sqrt{3},\sqrt{5},-\sqrt{5},\sqrt{7},-\sqrt{7},\ldots$ gives us an example where the $N_{\epsilon,j}$ are not bounded.On the other hand, it's easy to show that if $\lim_{n \to +\infty} a_k = a = \lim_{k \to +\infty} a_{-k}$ then the $N_{\epsilon,j}$ can be bound.

My intuition is that for the $N_{\epsilon,j}$ to explode when we start the average further into the sequence we need the terms in the sequence to be larger and larger.

I can't find a counter-example with this additional condition, nor can I use this condition to prove what I'm looking for.I've already asked this question to quite a few people and nobody seems to agree whether the statement I'm trying to demonstrate seems correct or not.

If anyone has an idea of a direction to take or finds a counter-example, it will be more than welcome.Thank you very much!

Originally, my question concerned the Birkhoff average for a smooth function $h$ on a closed manifold $M$ with a diffeomorphism $\varphi$.If the Birkhoff average exists at a point $x\in M$ is there a sufficiently large $N$ so that for all $j\in \mathbb{Z}$ and $n\geq N$, $\frac{1}{n}\sum_{k=1}^n h\circ \varphi^{k+j}(x)$ is close enough to the Birkhoff average.


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