Book of Proof - Amazon S3

156 Mathematical Induction
To round out this section, we present four additional induction proofs.
Proposition
If n ∈ Z and n ≥ 0, then
n∑
i · i! = (n + 1)! − 1.
Proof. We will prove this with mathematical induction.
(1) If n = 0, this statement is
i=0
0∑
i · i! = (0 + 1)! − 1.
i=0
Since the left-hand side is 0 · 0! = 0, and the right-hand side is 1! − 1 = 0,
the equation ∑ 0
i=0
i · i! = (0 + 1)! − 1 holds, as both sides are zero.
(2) Consider any integer k ≥ 0. We must show that S k implies S k+1 . That
is, we must show that
k∑
i · i! = (k + 1)! − 1
implies
k+1 ∑
i=0
i=0
We use direct proof. Suppose
k+1 ∑
i=0
i · i! =
=
i · i! = ((k + 1) + 1)! − 1.
k∑
i · i! = (k + 1)! − 1. Observe that
i=0
( ) k∑
i · i! + (k + 1)(k + 1)!
i=0
( )
(k + 1)! − 1 + (k + 1)(k + 1)!
= (k + 1)! + (k + 1)(k + 1)! − 1
= ( 1 + (k + 1) ) (k + 1)! − 1
= (k + 2)(k + 1)! − 1
= (k + 2)! − 1
= ((k + 1) + 1)! − 1.
Therefore
k+1 ∑
i=0
i · i! = ((k + 1) + 1)! − 1.
It follows by induction that
n∑
i · i! = (n + 1)! − 1 for every integer n ≥ 0.
i=0
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