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The formula for the Gordon growth model is:

$\hspace{1in}P= \sum_{t=1}^{\infty} D\times\frac{(1+g)^t}{(1+k)^t}$

So summing the infinite series we get:

$\hspace{1in}P=\frac{D(1+g)}{k-g} \hspace{2in}\mbox{(1)}$

Here's my attempt to arrive at equation (1):

The sum of a geometric series is:

$\hspace{1in}\frac{a}{1-r}$

So,plugging in the values:

$\hspace{1in}\frac{D}{1-\frac{1+g}{1+k}}$

$\hspace{1in}\frac{D}{\frac{1+k-(1+g)}{1+k}}$

$\hspace{1in}\frac{D}{\frac{k-g}{1+k}}$

Therefore,

$\hspace{1in}P=\frac{D(1+k)}{1+g}$

Appreciate if someone could highlight where I made a mistake.

Thanks you.

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You don't have the right value of a. –  Qiaochu Yuan Jan 13 '11 at 20:22
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What Qiaochu Yuan says. In fact, you probably just copied the formula for the sum from some book or internetpage and overlooked that the sum for that formula probably starts from $0$, not from $1$ like in your formula. –  Raskolnikov Jan 13 '11 at 20:24
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$\sum\nolimits_{t = 1}^\infty {x^t } = x\sum\nolimits_{t = 0}^\infty {x^t } = \frac{x}{{1 - x}}$, if $|x|<1$. So,... –  Shai Covo Jan 13 '11 at 23:08
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1 Answer

The question was basically answered already in comments. I'll add some details so that the question does not remain unanswered.

You were doing fine, but in the formula $\frac{a}{1-r}$ you should have used $$a=D\frac{1+g}{1+k} \text{ and }r=\frac{1+g}{1+k}.$$ (You have used $a=D$.)

With this change, using the same manipulation as you posted in your question, you will arrive to $$\frac a{1-r} = D\frac{1+g}{1+k} \frac1{1-\frac{1+g}{1+k}} = D\frac{1+g}{1+k} \frac1{\frac{k-g}{1+k}} = D\frac{1+g}{k-g}.$$

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