# Fraction of an exponential pyramid taken by all but the last row

Say I have a pyramid that grows by powers of 3:

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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?

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Hint 1: For all $m\geq 0$ and any $k\neq 1$, we have $$\sum_{i=0}^mk^i=\frac{k^{m+1}-1}{k-1}.$$
Last Step: For any $n\geq 0$ and any $k>1$, we have $$\frac{k^n-1}{k^{n+1}-1}=\cfrac{\frac1{k^n}\left(k^n-1\right)}{\frac1{k^n}\left(k^{n+1}-1\right)}=\cfrac{1-\frac1{k^n}}{k-\frac1{k^n}},$$ whence taking the limit as $n\to\infty$ yields the desired result.
So, in the example for $3$, that leaves us $\lim_{n \to \infty} \frac{3^n - 1}{3^{n+1} - 1} = ?$ – jamaicanworm Nov 28 '12 at 3:55
There you go!${}$ – Cameron Buie Nov 29 '12 at 4:56