I hate to disagree with Big Beach Banana, but it appears to me that you are actually using the Gordon growth model, which is one variant of the dividend discount model, which in turn is a variant of the capital asset pricing model. In the Gordon model, the stock price is not invariant.
Let’s first review a basic bit of math. If n is a positive integer and a is not equal to 1
[math](1 - a) * \sum_{j=0}^n a^j = \sum_{j=0}^n a^j - \left ( \sum_{j=0}^n a^{(j+1)} \right ) = \\
a^0 + \left ( \sum_{j=1}^n a^j \right ) - \left \{ \left ( \sum_{j=1}^n a^j \right ) + a^{(j+1)} \right \} = 1 - a^{(n+1)} \iff \\
\sum_{j=0}^n a^j = \dfrac{1 - a^{(n+1)}}{1 - a}.[/math]
Does that ring a bell? Geometric series?
[math]\therefore b > 0 \implies \sum_{j=1}^n \dfrac{1}{(1 + b)^j} = \sum_{j=1}^n \left ( \dfrac{1}{1 + b} \right )^j = \left \{ \sum_{j=0}^n \left ( \dfrac{1}{1 + b} \right )^j \right \} - 1=\\
\dfrac{1 - \left ( \dfrac{1}{1 + b} \right )^{(n+1)}}{1 - \dfrac{1}{1 + b}} - 1= \dfrac{1 - \dfrac{1}{(1 + b)^{(n+1)}} - \left ( 1 - \dfrac{1}{1 + b} \right )}{\dfrac{b}{1 + b}} =\\
\dfrac{\dfrac{1}{1 + b} - \dfrac{1}{(1 + b)^{(n+1)}}}{1} * \dfrac{(1 + b)}{b} = \dfrac{1 - (1 + b)^{-n}}{b}.[/math]
If b is an interest rate, this is just the formula for the present value an annuity of 1 dollar for n periods. This in turn gives us the present value of a perpetual annuity, namely [imath]1/b[/imath].
The capital asset pricing model basically says that the present value of an asset is the sum of its discounted cash flows. Applied to a stock that gives:
[math]V_t = \dfrac{(V_{t+1} + d_i * V_t)}{1 + r_i}, \text { where }\\
V_i = \text {market value of one share at start of period i } i;\\
d_i = \text {dividend amount payable at the end of period } i; \text { and}\\
r_i = \text {discount rate applicable to period } i.[/math]
The simplest assumption gives rise to the constant dividend discount model where the dividend is assumed certain, constant, and perpetual and the discount rate is assumed to be constant. Furthermore, at long term equilibrium, it must be true that
[math]V_t = V_{t+1} \implies V_{t+1} = \dfrac{V_{t+1} + d}{1 + r} \implies\\ V_{t+1}(1 + r) = V_{t+1} + d \implies V_{t+1} * r = d \implies r = \dfrac{d}{V_{t+1}} = \dfrac{d}{V_t}.[/math]
[math]\dfrac{d}{V_t} = r \implies V_t = \dfrac{d}{r}.[/math]
The Gordon model is not a long-term equilibrium model. The assumption is that
[math]d_i = d(1 + g)^i, \text { where } g > 0. [/math]
[math]V_t = \dfrac{V_{t+1} + d}{(1 + r)} \implies V_t = \dfrac{V_{t+n}}{(1 + r)^n} + d *\sum_{j=1}^n \left (\dfrac{1 + g}{1 + r} \right)^j[/math]
[math]V_0 = \dfrac{V_n}{(1 + r)^n} + d * \sum_{j=1}^n \left ( \dfrac{1 + g}{1+r} \right)^n.[/math]
I have to take a break now. I'll get back to it later
