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I have a question regarding infinite series. Is it true that \begin{equation} \frac{1}{n}\sum_{i=1}^n \cosh{\left(\frac{1}{3}\gamma^3x_i\right)}\underset{n\to\infty}\longrightarrow 1 \end{equation} if and only if \begin{equation}\frac{1}{n}\sum_{i=1}^n \vert x_i\vert \underset{n\to\infty}\longrightarrow 0,\end{equation} where $\gamma >0$ is a constant, and $\sup_i \vert x_i \vert < \infty$.

Any ideas (a proof?)??? Every help/hint is really much appreciated!

many thanks!

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    yes, I agree. I just stated the problem in the way that I encountered it. as it turns out (see below), it is really irrelevant. thanks for looking at it!2012-12-01

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Let $\alpha = \gamma^3/3$. The following inequality holds for the convex function $x \mapsto \cosh(\alpha x)$:

$ \sum_{k=1}^n \frac{\cosh(\alpha x_k)}{n} = \sum_{k=1}^n \frac{\cosh(\alpha |x_k|)}{n}\geq \cosh \left(\alpha \sum_{k=1}^n \frac{|x_k|}{n}\right) \geq 1. $

(The mean of some function values is at least the function value of the mean.) If the first term converges to $1$ then so does the last $\cosh$ term. This shows that $\sum_{k=1}^n \frac{|x_k|}{n}$ converges to $0$. If $|x| \leq N$ then $\cosh(\alpha x) \leq 1 + |x| \frac{\cosh(\alpha N) - 1}{N}.$ The reverse implication follows directly from this inequality since $|x_k| \leq N$ for some bound $N$ and all indices $k$.

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    thank you so much, that's a great proof! I really appreciate your support in this!2012-12-01