# convergence of a series involving $\cosh$

I have a question regarding infinite series. Is it true that $$\frac{1}{n}\sum_{i=1}^n \cosh{\left(\frac{1}{3}\gamma^3x_i\right)}\underset{n\to\infty}\longrightarrow 1$$ if and only if $$\frac{1}{n}\sum_{i=1}^n \vert x_i\vert \underset{n\to\infty}\longrightarrow 0,$$ 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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That $\gamma^3/3$ there by the way is really irrelevant to the problem. You could just as well assume that it's $1$. It would simplify the expressions (I don't think it helps much though, it's just an opinion)! – Patrick Da Silva Dec 1 '12 at 9:18
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! – s_2 Dec 1 '12 at 13:32

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 $k$.