Assuming that the annual interest rate $r$ is constant, show that the present value of an infinite stream of annual payments of the form $C,Cg,Cg^2,\dots$ growing at a constant rate g, is given by the formula $$\frac{C}{1+r-g}$$ I don't understand at all how to get this result. I see that $$a+ar+ar^2+\cdots = \sum_{k=0}^{\infty}ar^k = \frac{a}{1-r}$$ So this would be $$C+Cg+Cg^2+\cdots = \sum_{k=0}^{\infty}Cg^k = \frac{C}{1-g}$$ How am I supposed to have the result above? I don't see how at all.
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Your last series expression is not the present value. It does not include discounting. What would be the present value of an infinite stream of constant annual payments $C, C, \dots$w.r.t. a discount rate $r$? – Roland Feb 03 '18 at 18:19
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I don't know... Like what is even meant by the present value? What would that be? Would it be $C+Cr+Cr^2+\cdots$ ?? @Roland – MRT Feb 03 '18 at 18:21
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If $\left|\frac{g}{1+r}\right|<1$ we have
$$ \begin{align} PV &= \frac{C}{1+r}+\frac{Cg}{(1+r)^2}+\frac{Cg^2}{(1+r)^3}+\cdots\\ &=\frac{C}{g}\times\sum_{k=1}^\infty \frac{g^k}{(1+r)^k}\\ &=\frac{C}{g}\times \frac{\frac{g}{1+r}}{1-\frac{g}{1+r}}\\ &=\frac{C}{1+r-g} \end{align} $$ using $\sum_{k=1}^\infty x^k=\frac{x}{1-x}$ for $|x|<1$.
alexjo
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Consider
$$C+C(g-r)+C(g-r)^2+.....=\frac {C}{1+r-g}.$$
Note that the rate in this case is rate in minus rate out .
Mohammad Riazi-Kermani
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