How did you get from [b1(a1 - d)]*rn-1 to [b1(a1 - d)] / (1 - r) ???
Hi CPhill,
first i tell you about der geometric sequence of beta:
$$\small{\text{
The geometric sequence is
$
b_1 = \underbrace{9*(\frac{2}{3})^0}_{=9}, \ b_2 = \underbrace{9*(\frac{2}{3})^1}_{=6}, \ b_3 = \underbrace{9*(\frac{2}{3})^2}_{=4}, \ b_4 = \underbrace{9*(\frac{2}{3})^3}_{=\frac{8}{3}}, \ \dots \ ,\ b_n = 9*(\frac{2}{3})^{n-1}
$
}} \\
\small{\text{
The sum $s=\frac{9}{1-\frac{2}{3}}$\ , now we set $\ b_1 = 9 $ and $\ r = \frac{2}{3}
$ and see why
}}\\\\
\small{\text{
b_1 = b_1, \ b_2 = b_1 r, \ b_3 = b_1r^2, \ b_4 = b_1r^3, \ \dots \ ,\ b_n = b_1r^{n-1}
$
}} \\
\small{\text{
The sum is
$
s = b_1 + b_1 r + b_1r^2 + b_1*r^3 + b_1r^4 + \ \dots
$
}} \\
\small{\text{
$
\begin{array}{rcll}
\hline
s &=& b_1 \ + & b_1 r + b_1r^2 + b_1*r^3 + b_1r^4 + \ \dots \\
rs &=& & b_1 r + b_1r^2 + b_1*r^3 + b_1r^4 + \ \dots \\
\hline
s-rs &=& b_1 & + 0 + 0 + 0 + 0 +\dots \\
s(1-r) &=& b_1 & \\
s &=& \dfrac{b_1}{1-r} &
\end{array}
$
}} \\
\boxed{
\small{\text{
So, if we have a Geometric Sequence $ b_n = b_1r^{n-1} $ the infinite sum $ s = \dfrac{b_1}{1-r} $
}}
}$$
Now we look to Gamma Sequence. There is a constant: $$b_1(a_1-d)=g_1$$:
$$\boxed{
\small{\text{
The Geometric Sequence Gamma first part is $ g_{n_{First Part}} = g_1r^{n-1} $ the infinite sum $ s = \dfrac{g_1}{1-r} $
}}
}\\\\
\small{\text{
$ s = \dfrac{g_1}{1-r} =\dfrac{b_1(a_1-d)}{1-r}$
}}\\\\
\small{\text{
set $ b_1 = 9, \ a_1 = 10,\ d = -2, \ r =\frac{2}{3}
$
}}\\
\small{\text{
we have the sum of Gamma Sequence first part:
}}\\
\small{\text{
$
108*(\frac{2}{3})^0 + 108*(\frac{2}{3})^1 + 108*(\frac{2}{3})^2 +108*(\frac{2}{3})^3 + \dots
$
}} \\
\small{\text{
and the infinite sum $s = \frac{108}{1-\frac{2}{3}}$
and $g_{1_{first part}}=108,\ g_{2_{first part}}=108*\frac{2}{3}, \ \dots \ ,\ g_{n_{first part}}=108*(\frac{2}{3})^{n-1} $
}}\\$$
P.S.
$$\\\small{\text{
The finite sum:
$
s_n=a_1 + a_1\cdot r + a_1\cdot r^2 + a_1\cdot r^3 + \dots +a_1\cdot r^{n-1}
$
is
$
s_n = \dfrac{a_1(1-r^n)}{1-r} $
}}\\\\
\small{\text{
The infinite sum:
$
s_n=a_1 + a_1\cdot r + a_1\cdot r^2 + a_1\cdot r^3 +a_1\cdot r^4 \dots
$
}}\\\\
\small{\text{
with
$
s_n = \dfrac{a_1(1-r^n)}{1-r} $
is $s=\lim\limits_{n\to \infty} s_n = \dfrac{a_1}{1-r}$,}}\\\\
\small{\text{
if $\quad -1 }}$$
.
