Say I have a pyramid that grows by powers of 3:
* *** *********
I'd like to know what fraction of the pyramid is taken by every row except the last. In the above picture, it would be $\frac{1+3}{1+3+9} = 0.31$; but what happens as the number of rows increases?
Let
$s_n = \frac{\sum_{i=0}^{n-1} 3^i}{\sum_{i=0}^n 3^i}.$
We can see that the first few terms of the series are
$s_1 = \frac{1}{1+3} = 0.25$ $s_2 = \frac{1+3}{1+3+9} = 0.31$ $s_3 = \frac{1+3+9}{1+3+9+27} = 0.325$ $\vdots$
Is it true that $s_n \to \frac{1}{3}$ as $n \to \infty$? Can this be proven?
Computationally, it also seems like $\lim_{n \to \infty} \frac{\sum_{i=0}^{n-1} 2^i}{\sum_{i=0}^n 2^i} = \frac{1}{2}.$
In general, is it true that the fraction of a pyramid (growing exponentially by $k$) occupied by all but the last row tends to $\frac{1}{k}$ as the pyramid grows?