Show that for g < r, [ sum_{k=1}^{infty} frac{(1+g)^{k-1}}{(1+r)^{k}}=frac{1}{r-g} ] [Let $S$ be the value of the
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Show that for g < r,
\[ \sum_{k=1}^{\infty} \frac{(1+g)^{k-1}}{(1+r)^{k}}=\frac{1}{r-g} \]
[Let $S$ be the value of the sum. Note that $S=1 /(1+r)+S(1+g) /(1+r)$.]
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