I have a periodic progression $S=\{a,b,c,d,a,b,c,d,\ldots,a,b,c,d\}$ and I compute its mean as function of the elements cover i.e.
$\displaystyle A_j=\frac{1}{j}\sum_{i=1}^{j}S_i$, then $A_1=a,A_2=\frac{a+b}{2},\ldots$. I know in the $\displaystyle\lim_{j\rightarrow\infty}$ $A_j=\frac{a+b+c+d}{4}$ which is the average of $S$ in one period. How do I put down the rigorous proof of this results ?
Note. I got the result by computing the series by programing
Suppose you have $m$ items repeated periodically, $(a_k)_{i=1}^k $. Let $A =\frac1{m}\sum_{h=1}^m a_h $.
Then to get the average of the first $n$ items, write $n = im+j$ where $0 \le j \lt m$.
Then
$\begin{array}\\ \frac1{n}\sum_{h=1}^n a_h &=\frac1{im+j}\sum_{h=1}^{im+j} a_h\\ &=\frac1{im+j}\left(\sum_{h=1}^{im} a_h+\sum_{h=im+1}^{im+j} a_h\right) \qquad\text{split off beyond the multiples of } m\\ &=\frac1{im+j}\left(i\sum_{h=1}^{m} a_h+\sum_{h=im+1}^{im+j} a_h\right)\\ &=\frac1{im+j}\left(imA+\sum_{h=im+1}^{im+j} a_h\right) \qquad\text{since there are }i\text{ copies of the } m \text{ values}\\ &=\frac1{im+j}imA+\frac1{im+j}\sum_{h=im+1}^{im+j} a_h\\ &=\frac{im}{im+j}A+\frac1{im+j}\sum_{h=1}^{j} a_h \qquad\text{since the }a_h \text{ repeat}\\ &=\frac{im+j-j}{im+j}A+\frac1{im+j}\sum_{h=1}^{j} a_h \qquad\text{standard technique to get } A \text{ by itself}\\ &=(1-\frac{j}{im+j})A+\frac1{im+j}\sum_{h=1}^{j} a_h\\ &=A-\frac{j}{im+j}A+\frac1{im+j}\sum_{h=1}^{j} a_h\\ &=A-\frac1{im+j}(jA+\sum_{h=1}^{j} a_h)\\ &=A-\frac1{n}(jA-\sum_{h=1}^{j} a_h) \qquad\text{had "+" inside}\\ \end{array} $
Therefore
$\begin{array}\\ |\frac1{n}\sum_{h=1}^n a_h-A| &=|\frac1{n}(jA-\sum_{h=1}^{j} a_h)|\\ &\le\frac1{n}(m|A|+\sum_{h=1}^{m} |a_h|)\\ \end{array} $
Since all terms inside of the final absolute values are bounded, $\lim_{n \to \infty} |\frac1{n}\sum_{h=1}^n a_h-A| = 0$ or $\lim_{n \to \infty} \frac1{n}\sum_{h=1}^n a_h = A$.