Appendix A — Geometric Series

A geometric series is a sum of terms, where the ratio of each term to its predecessor is a constant. \[S_n = a+ar+ar^2 + ... + ar^{n-1} = \sum_{k=0}^{n-1} ar^j\] This sum is equal to: \[S_n = \frac{a(1-r^n)}{1-r}\] If \(-1 < r < 1\), then \[\lim_{n\to\infty} S_n = \frac{a}{1-r}\] Able to sum geometric series is very important, and it is crucial that we do not go into off-by-one errors.

Example: Sum the following series: * \(v^{10} + v^{15} + . . . + v^{65} = ?\) * \((1+i)^{44} +(1+i)^{40}(1+r)^{2} +(1+i)^{36}(1+r)^{4} +...+(1+i)^{4}(1+r)^{20} =?\)

Solution: \[v^{10} + v^{15} + . . . + v^{65} = v^{10} (1 + v^5 + ... + v^{55}) = v^{10} \frac{1-v^{60}}{1-v^5}\] \[\begin{align}&(1+i)^{44} +(1+i)^{40}(1+r)^{2} +(1+i)^{36}(1+r)^{4} +...+(1+i)^{4}(1+r)^{20} \\ &= (1+i)^{44} \big[1 + \frac{(1+r)^2}{(1+i)^4} + \frac{(1+r)^4}{(1+i)^8} + ... + \frac{(1+r)^{20}}{(1+i)^{40}}\big] \\ &= (1+i)^{44} \frac{1-(1+r)^{22}(1+i)^{-44}}{1-(1+r)^2(1+i)^{-4}} \end{align}\]

Example: Calculate the following series: \[v + 2v^2 + 3v^3 + 4v^4 + . . . + 20v^{20} = ?\] Solution: This is not a geometric series. However, we could use a trick to calculate it. Let \(A = v + 2v^2 + 3v^3 + 4v^4 + . . . + 20v^{20}\). Then \[\begin{align}(1+i)A &= 1&+ 2v + 3v^2 + 4v^3 + ... &+ 20 v^{19} \\ A &= &v + 2v^2 + 3v^3 + ... &+ 19v^{19} + 20v^{20}\end{align}\] Subtracting \((1+i)A - A\), we obtain \(iA = 1 + (v+v^2+...+v^{19}) + 20v^{20}\) Thus, \[A = \frac{1 + (v+v^2+...+v^{19}) + 20v^{20}}{i}\] And we will denote the series \(v+v^2+...+v^{19} = a_{\overline{19|}}\), the present value of an annuity-immediate.