Arithmetico-geometric sequence

Series

Partial sums

The sum of the first n terms of an arithmetico-geometric series has the form

${\displaystyle \begin{array}{cc}{S}_n& =\sum _{k=1}^n{A}_k{G}_k\\ & =\sum _{k=1}^n(a+(k-1)d)b{r}^{k-1}\\ & =b\sum _{k=0}^{n-1}\left(a+kd\right)r^k\\ & =ab+(a+d)br+(a+2d)b{r}^2+\cdots +(a+(n-1)d)b{r}^{n-1}\end{array}}$

where $A_i$ and $G_i$ are the ith elements of the arithmetic and the geometric sequence, respectively.

This partial sum has the closed-form expression

${\displaystyle \begin{array}{cc}{S}_n& =\frac{ab-(a+nd)b{r}^n}{1-r}+\frac{dbr(1-r^n)}{(1-r{)}^2}\\ & =\frac{A_1{G}_1-A_{n+1}{G}_{n+1}}{1-r}+\frac{dr}{(1-r{)}^2}(G_1-G_{n+1}).\end{array}}$

Derivation

Multiplying

${\displaystyle S_n=ab+(a+d)br+(a+2d)b{r}^2+\cdots +(a+(n-1)d)b{r}^{n-1}}$

by r gives

${\displaystyle r{S}_n=abr+(a+d)b{r}^2+(a+2d)b{r}^3+\cdots +(a+(n-1)d)b{r}^n.}$

Subtracting rS_n from S_n, dividing both sides by ${\displaystyle b}$, and using the technique of telescoping series (second equality) and the formula for the sum of a finite geometric series (fifth equality) gives

${\displaystyle \begin{array}{cc}\frac{(1-r)S_n}{b}& =\left(a+(a+d)r+(a+2d)r^2+\cdots +(a+(n-1)d)r^{n-1}\right)-(ar+(a+d)r^2+(a+2d)r^3+\cdots +(a+(n-1)d)r^n)\\ & =a+d\left(r+r^2+\cdots +r^{n-1}\right)-(a+(n-1)d)r^n\\ & =a+d\left(r+r^2+\cdots +r^{n-1}+r^n\right)-\left(a+nd\right)r^n\\ & =a+dr\left(1+r+r^2+\cdots +r^{n-1}\right)-\left(a+nd\right)r^n\\ & =a+\frac{dr(1-r^n)}{1-r}-(a+nd)r^n,\\ S_n& =\frac{b}{1-r}\left(a-(a+nd)r^n+\frac{dr(1-r^n)}{1-r}\right)\\ & =\frac{ab-(a+nd)b{r}^n}{1-r}+\frac{dr(b-b{r}^n)}{(1-r{)}^2}\\ & =\frac{A_1{G}_1-A_{n+1}{G}_{n+1}}{1-r}+\frac{dr(G_1-G_{n+1})}{(1-r{)}^2}\end{array}}$

as claimed.

Infinite series

If −1 < r < 1, then the sum S of the arithmetico-geometric series, that is to say, the limit of the partial sums of the elements of the sequence, is given by

${\displaystyle \begin{array}{cc}S& =\sum _{k=1}^{\mathrm{\infty }}{t}_k=\underset{n\to \mathrm{\infty }}{lim}{S}_n\\ & =\frac{ab}{1-r}+\frac{dbr}{(1-r{)}^2}\\ & =\frac{A_1{G}_1}{1-r}+\frac{dr{G}_1}{(1-r{)}^2}.\end{array}}$

If r is outside of the above range, b is not zero, and a and d are not both zero, the limit does not exist and the series is divergent.

Example

The sum

${\displaystyle S=\frac{0}{1}+\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\cdots }$,

is the sum of an arithmetico-geometric series defined by ${\displaystyle d=b=1}$, ${\displaystyle a=0}$, and ${\displaystyle r=\frac{1}{2}}$, and it converges to ${\displaystyle S=2}$. This sequence corresponds to the expected number of coin tosses required to obtain "tails". The probability ${\displaystyle T_k}$ of obtaining tails for the first time at the kth toss is as follows:

${\displaystyle T_1=\frac{1}{2},T_2=\frac{1}{4},\dots ,T_k=\frac{1}{2^k}}$.

Therefore, the expected number of tosses to reach the first "tails" is given by

${\displaystyle \sum _{k=1}^{\mathrm{\infty }}k{T}_k=\sum _{k=1}^{\mathrm{\infty }}\frac{k}{2^k}=2.}$

Similarly, the sum

${\displaystyle S=\frac{0\cdot \frac{1}{6}}{\frac{5}{6}}+\frac{1\cdot \frac{1}{6}}{1}+\frac{2\cdot \frac{1}{6}}{\frac{6}{5}}+\frac{3\cdot \frac{1}{6}}{{\left(\frac{6}{5}\right)}^2}+\frac{4\cdot \frac{1}{6}}{{\left(\frac{6}{5}\right)}^3}+\frac{5\cdot \frac{1}{6}}{{\left(\frac{6}{5}\right)}^4}+\cdots }$

is the sum of an arithmetico-geometric series defined by ${\displaystyle d=1}$, ${\displaystyle a=0}$, ${\displaystyle b=\frac{1/6}{5/6}=\frac{1}{5}}$, and ${\displaystyle r=\frac{5}{6}}$, and it converges to 6. This sequence corresponds to the expected number of six-sided dice rolls required to obtain a specific value on a die roll, for instance "5". In general, these series with ${\displaystyle d=1}$, ${\displaystyle a=0}$, ${\displaystyle b=\frac{p}{1-p}}$, and ${\displaystyle r=1-p}$ give the expectations of "the number of trials until first success" in Bernoulli processes with "success probability" ${\displaystyle p}$. The probabilities of each outcome follow a geometric distribution and provide the geometric sequence factors in the terms of the series, while the number of trials per outcome provides the arithmetic sequence factors in the terms.