GEOMETRIC SEQUENCES (GEOMETRIC PROGRESSIONS)
Definition
A sequence a1
,
a2
,
a3
, K
an
,
is a geometric sequence (or geometric progression) if
the quotient of any two successive terms of the sequence is a constant.
a
The constant ratio of any two consecutive terms n = r is called common ratio
a n −1
n ≥ 2 , n ∈ Z , r ≠ 1, r ≠ 0 .
Nth Term of a Geometric Sequence
a1
a2
= a1 r
a3
= a1 r 2
a4
= a1 r 3
M
an
= a1 r n −1
Sum of the First n Terms of an Geometric Sequence (Progression)
For the geometric sequence (progression) a1 , a 2 , a3 , K a n ,
an
=r
a n −1
Define the partial sums:
S1 = a1
S 2 = a1 + a 2 = a1 + a1 r = a1 (1 + r )
S 3 = a1 + a 2 + a3 = a1 + a1 r + a1 r 2 = a1 (1 + r + r 2 )
M
S n = a1 + a 2 + a3 + L + a n = a1 + a1 r + a1 r 2 + L + a1 r n −1 = a1 (1 + r + r 2 + L + r n −1 )
S n = a1 + a1 r + a1 r 2 + L + a1 r n −1
rS n = r (a1 + a1 r + a1 r 2 + L + a1 r n −1 ) = a1 + a1 r 2 + a1 r 3 + L + a1 r n
Subtract
S n − rS n = a1 + a1 r + a1 r 2 + L + a1 r n −1 − ( a1 + a1 r 2 + a1 r 3 + L + a1 r n )
S n (1 − r ) = a1 + a1 r + a1 r 2 + L + a1 r n −1 − a1 − a1 r 2 − a1 r 3 − L − a1 r n = a1 (1 − r n )
S n = a1
1− rn
1− r
n
n
n
k =1
k =1
k =1
Using sigma (summation) notation S n = ∑ a k =∑ a1 r k −1 = a1 ∑ r k −1 = a1
1− rn
1− r
NOTE
1+ r + r +L + r
2
n −1
1− rn
=
1− r
(e.g.) Chess legend (from G. Gamow:”1, 2, 3,…. ∞ ”)
According to legend, the game of chess was invented by Grand Vizier Sissa Ben Dahir, who presented it to
King Shirham of India as a gift. In gratitude, the king offered the Grand Visir any reward requested -provided, of course, that it sounded reasonable. The Grand Vizier asked for only:
"One grain of wheat, representing the first square of a chessboard. Two grains for the
second square. Four grains for the next. Then eight, sixteen, thirty two … doubling
for each successive square until the 64th and last square is counted."
The king was impressed with the apparent modesty of the request, and he immediately granted it. He took
his chessboard, removed the pieces, and asked for a bag of wheat to be brought in. But, to his surprise, the
bag was emptied by the 20th square. The king had another bag brought in, but then realized that the entire
second bag was needed for just the next square. In fact, in 20 more squares, he would need as many bags as
there were grains of wheat in the first bag! And that was only up to square 40. Legend does not record
what the king then did to his Grand Vizier.
Note: If we work out the number of grains of wheat which Sissa Ben Dahir was to receive, we would add,
1 + 2 + 2 2 + 2 3 + ... + 2 64 . We would do this with the formula for a geometric series,
a (r n − 1)
1(2 64 − 1)
and get, S 64 =
= 18,446,744,073,709,551,615 grains of wheat.
Sn =
r −1
2 −1
A conservative estimate is that a bushel of wheat contains 5 million grains. This means to fulfill the
reward, it would take about 4,000 billion bushels. The estimated production of wheat in those days
averaged 2 billion bushels a year thus the amount requested by Sissa Ben Dahir was all of the world’s
production of wheat for the next 2000 years.
a1 = 1 , a 2 = 2 , a3 = 4 , a 4 = 8L, a 64 = 2 63
S 64 = 1 + 2 + 2 2 + 2 2 + L 2 63 = 1 ⋅
1 − 2 64
= 18,446,744,073,709,551,615
1− 2
Infinite Geometric Series
Let S ∞ = lim S n
n →∞
S ∞ = a1 + a 2 + a3 + L + a n + L = a1 + a1 r + a1 r 2 + L + a1 r n −1 + L = a1 (1 + r + r 2 + L + r n −1 + L)
1− rn
n →∞ 1 − r
S ∞ = lim S n = a1 lim
n →∞
if | r |< 1 lim r n = 0 and S ∞ = a1
n→∞
1
1− r
if | r | ≥ 1 lim r n does not exist
n →∞
S∞ =
a1
1− r
only if | r | < 1
(e. g.) Zeno’s first paradox
1
1
1
1
1
a1 = , a 2 = , a3 = , a 4 = L, a n = n ,L is an infinite geometric progression
2
4
8
16
2
1
(sequence) with common ratio r =
2
Let’s now compute the infinite series
1 1
1
1
1
+ 2 + 3 + 4 +L+ n +L
2 2
2
2
2
1
1 − ( )n
1 1
1
1
1
1
2
Sn = + 2 + 3 + 4 + L + n =
1
2 2
2
2
2
2
1−
2
S∞ =
a1
=
1− r
1
2
1−
1
2
=1
⇒ S∞ =
1
2
1−
1
2