Formulated a Series Problem But Unfortunately Don't Know How To Solve

This all started when I was playing around with a financial spreadsheet. There is no need to know financial terms as I've managed to convert this observation into a mathematical problem. But just in case someone knows it was simply the observation that if you have a constant growth rate in Net Income and reinvest everything soon the Return on Equity will approach the same constant growth rate. Let Net Income at time $$0$$ be defined as $$N_0$$ and the Book Value of Equity at time $$0$$ be defined as $$B_0$$. Then Net Income at time $$n$$ is defined as $$N_n=N_0(1+g)^n$$ and $$B$$ at time $$n$$ is defined as $$B_n = B_0+\sum_{t=1}^{n}N_0(1+g)^n$$. My observation was that the fraction $$\frac{N_n}{B_{n-1}}=\frac{N_0(1+g)^n}{B_0+\sum_{t=1}^{n-1}N_0(1+g)^{t}}\rightarrow g$$ after some time $$n$$. I'm almost certain I'm complicating this but can someone help me figure how long it will take, $$n$$, this to converge if $$N_0$$, $$B_0$$, and $$g$$ are given. Or will this be the limit as $$n\rightarrow \infty$$?

• Your observation is correct. To prove it, you can sum the geometric series in the denominator and use algebra to simplify the result. There will be terms like $(1+g)^{-n}$ which go to zero, and $N_0$ cancels. The result will indeed be $g$. – Hans Engler Dec 9 '18 at 23:50
• Thanks @HansEngler, I have a quick follow up question. Suppose $S_n=\frac{N_n}{B_{n-1}}$ then each ratio is $\frac{S_{n+1}}{S_n} - 1$ away from $g$. Why is that a case or can you point me in the right direction to see. Thanks again. – Dmitriy Dec 11 '18 at 3:28

\begin{align}\frac{N_0(1+g)^n}{B_0 + \sum_{t=1}^{n-1}N_0 (1+g)^t}&= \frac{N_0(1+g)^n}{B_0 + \frac{N_0 (1+g)[(1+g)^{n-1}-1]}{g}}\\ &= \frac{N_0(1+g)^ng}{B_0g + N_0 [(1+g)^{n}-(1+g)]}\\ &=g\left(\frac{1}{1+\frac{B_0g-(1+g)}{N_0(1+g)^n}}\right)\end{align}
Hence, as $$n \to \infty$$, We have
\begin{align}\lim_{n \to \infty}\frac{N_0(1+g)^n}{B_0 + \sum_{t=1}^{n-1}N_0 (1+g)^t} =\lim_{n \to \infty}g\left(\frac{1}{1+\frac{B_0g-(1+g)}{N_0(1+g)^n}}\right)=g\left( \frac{1}{1+0} \right)=g\end{align}