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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/sn−1. The sequence {rn}

itself grows rapidly. Examining the ratios of ratios, a familiar pattern emerges:

n snrn=sn/sn−1≈rn/rn−1≈

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/sn−1=e.

Proof. By direct calculation we get

sn=(n!)n+1

n

Y

k=0

(k!)−2,n≥0

sn/sn−1=nn

n!,n≥1,and

sn+1/sn

sn/sn−1=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/sn−1)=

(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