Curious limit of a sequence used to prove Etemadi's SLLN

Solution 1:

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 $$ \lim_{k\to\infty} \frac1k\sum_1^{k}x_i = c.$$