Sum dissapearing when we assume some elements to be constant over time

Question

I have the dividend discount model, which is the following expression:

$$ P_{j,t} = \sum_{\tau=1}^{\infty}D_\tau(1+g)^\tau(1+r)^{-\tau}=\frac{D_{\tau+1}}{r-g} $$

Where $D_{\tau}$, is the dividend at time $t$, $g$ represents the constant growth over time and $r$ represents the required rate of return which is assumed to be constant over time too.

My questions are: why under constant growth and constant required rate of return can we re-write it this way. Why does the sum sign disappears?

Thank you.

Answer 1

You have $D_\tau$ in the sum, but it should probably be $D_0$ or $D_t$ (should not depend on the summation index, as the growing dividends are already accounted for in the sum with the growth rate).

Assuming it is $D_0$, the sum, which we shall call $S$, is $$S=\sum\limits_{\tau=1}^{\infty}D_0(1+g)^\tau(1+r)^{-\tau} =D_0\sum\limits_{\tau=1}^{\infty}\left(\frac{1+g}{1+r}\right)^{\tau}.$$ The reason we can simplify this is due to the geometric series formula, which is a formula in mathematics that tells us that if $c$ is a constant with $\color{red}{|c|<1}$, then

$$\sum\limits_{\tau=1}^{\infty}c^{\tau} = \frac{c}{1-c}.$$

Applying this formula with our constant $c=\frac{1+g}{1+r}$ (which satisfies $|c|<1$ if $0\le g < r$ for example), we see that $$\begin{align} S &= D_0\sum\limits_{\tau=1}^{\infty}\left(\frac{1+g}{1+r}\right)^{\tau} \\ &= D_0 \times \frac{\frac{1+g}{1+r}}{1-\frac{1+g}{1+r}}\\ &= D_0\times \frac{1+g}{r-g} \tag{simplifying the fraction}\\ &= \frac{D_1}{r-g}, \end{align}$$ since $D_1 = D_0(1+g)$. If you want $D_t$ instead of $D_0$, the $D_1$ at the end becomes $D_t (1+g)=D_{t+1}$.

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In summary, the reason for your formula is the geometric series formula

$$ \sum\limits _{\tau=1}^{\infty}c^\tau =\frac{c}{1-c},$$

which is true for any constant $c$ with $|c| < 1$.