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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 $n\rightarrow \infty$?

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    $\begingroup$ 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 $g$. $\endgroup$ – Hans Engler Dec 9 '18 at 23:50
  • $\begingroup$ 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. $\endgroup$ – Dmitriy Dec 11 '18 at 3:28
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\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}

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