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Friday, September 26, 2008

Derivation

Derivation

We want to find out the value of Pn as  n \rightarrow \infty , where

P_n = \sum_{t=1}^{n}  D\times \frac{(1+g)^t}{(1+k)^t} = D\times \sum_{t=1}^{n} \frac{(1+g)^t}{(1+k)^t}  = D\times \sum_{t=1}^{n} \left(\frac{1+g}{1+k}\right)^t

Let

 a = \frac{1+g}{1+k} .

Then

P_n = D\times \sum_{t=1}^{n} a^t .

Since

 1-a^{(n+1)} = (1-a) \times (1 + a + a^2 + ... + a^n) = (1-a) \times (1 + \sum_{t=1}^{n} a^t) ,

we get

 \frac{1-a^{(n+1)}}{1-a} - 1 = \sum_{t=1}^{n} a^t .

Therefore,

P_n = D \times \left[ \frac{1-a^{(n+1)}}{1-a} - 1 \right] .

If g < k, then a <> and

 a^{(n+1)} \rightarrow 0 as  n \rightarrow \infty .

Thus, we get

P_\infty = D \times \left( \frac{1}{1-a} - 1 \right) = D \times \left( \frac{a}{1-a} \right) = D \times \left(\frac{1+g}{1+k}\right) / \left[ 1 - \left(\frac{1+g}{1+k}\right) \right] = D \times \left( \frac{1+g}{k-g} \right) .

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