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# Math Bite: Finding e in Pascal’s Triangle

Authors:
• Brothers Technology

## Abstract

If s_n is the product of the entries in row n of Pascal’s triangle then (s_(n + 1)/s_n)/(s_n/s_(n - 1)) = (1 + 1/n)^n, which has the limiting value e.
VOL. 85, NO. 1, FEBRUARY 2012 51
Math Bite: Finding ein Pascal’s Triangle
HARLAN J. BROTHERS
Brothers Technology, LLC
Branford, CT 06405
harlan@brotherstechnology.com
Mathematicians have long been familiar with the tidy way in which the nth row of
Pascal’s triangle sums to 2n(the top row conventionally labeled as n=0). It is less
obvious how the rows behave when we multiply their items.
1 ................1
1 1 ................1
1 2 1 ................2
1 3 3 1 ................9
1 4 6 4 1 ...............96
1 5 10 10 5 1 . . . . . . . . . . . . 2500
1 6 15 20 15 6 1 . . . . . . . . . . 162000
Let snbe the product for row n; that is, sn=Qn
k=0n
k. On the right-hand side of the
ﬁgure above, we see the sequence {sn}grows very quickly. To get a sense of its rate of
growth, we can look at the ratios of successive terms, rn=sn/sn1. The sequence {rn}
itself grows rapidly. Examining the ratios of ratios, a familiar pattern emerges:
n snrn=sn/sn1rn/rn1
1 1 1
2 2 2 2
3 9 4.5 2.25
4 96 10.667 2.370
5 2500 26.042 2.441
6 162000 64.800 2.488
.
.
..
.
..
.
..
.
.
1000 1.68 10215681 2.49 10432 2.71692
THE OREM. lim
n→∞
sn+1/sn
sn/sn1=e.
Proof. By direct calculation we get
sn=(n!)n+1
n
Y
k=0
(k!)2,n0
sn/sn1=nn
n!,n1,and
sn+1/sn
sn/sn1=1+1
nn
.
Given that lim
n→∞ 1+1
nn
=e, the result follows.
Summary If snis the product of the entries in row nof Pascal’s triangle then (sn+1/sn)/(sn/sn1)=
(1+1/n)n, which has the limiting value e.
Math. Mag. 85 (2012) 51.doi:10.4169/math.mag.85.1.51.c
Mathematical Association of America
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