MATH1050 Bernoulli’s Inequality.

1. Recall the results Theorem (⋆), Theorem (⋆⋆) from the handout Arithmetic progression and geometric progression.
Theorem (⋆).
Suppose n ∈ N and r ∈ C. Then the statements below hold:

(a) 1 − rn+1 = (1 − r)(1 + r + r2 + · · · + rn ).

1 − rn+1
(b) Further suppose r ̸= 1. Then = 1 + r + r2 + · · · + rn .
1−r
Theorem (⋆⋆).
Suppose n ∈ N, and s, t ∈ C.
Then the equality sn+1 − tn+1 = (s − t)(sn + sn−1 t + sn−2 t2 + · · · + sn−k tk + · · · + stn−1 + tn ) holds.

Assuming the validity of (⋆), we deduce (⋆⋆) below. We will then apply (⋆⋆) to deduce the result known as ‘Bernoulli’s
Inequality’.

2. Proof of Theorem (⋆⋆).

Suppose n ∈ N, and s, t ∈ C.

• (Case 1.) Suppose s = t = 0. Then sn+1 −tn+1 = 0 = (s−t)(sn +sn−1 t+sn−2 t2 +· · ·+sn−k tk +· · ·+stn−1 +tn ).
• (Case 2.) Suppose s ̸= 0. Then t/s is well-defined as a number. Write r = t/s.
Then we have
sn+1 − tn+1 = sn+1 (1 − rn+1 )
= sn+1 (1 − r)(1 + r + r2 + · · · + rk + · · · + rn−1 + rn )
= (s − t)(sn + sn−1 t + sn−2 t2 + · · · + sn−k tk + · · · + stn−1 + tn )

• (Case 3.) Suppose t ̸= 0. By modifying the argument in Case 2, (interchanging the respective roles of s, t), we
also deduce the equality

sn+1 − tn+1 = (s − t)(sn + sn−1 t + sn−2 t2 + · · · + sn−k tk + · · · + stn−1 + tn ).

Hence, in any case, sn+1 − tn+1 = (s − t)(sn + sn−1 t + sn−2 t2 + · · · + sn−k tk + · · · + stn−1 + tn ).

3. Theorem (1). (Bernoulli’s Inequality.)

Let m ∈ N\{0, 1} and β ∈ R. Suppose β > −1. Then (1 + β)m ≥ 1 + mβ. Equality holds iff β = 0.
Proof of Theorem (1).
Let m ∈ N\{0, 1} and β ∈ R. Suppose β > −1.
[Preparatory step.] Note that

(1 + β)m − 1 = (1 + β)m − 1m = [(1 + β) − 1][(1 + β)m−1 + (1 + β)m−2 + · · · + 1]

= β[(1 + β)m−1 + (1 + β)m−2 + · · · + (1 + β) + 1]

[Reminder. We now deduce the statements below:

(1) If β > 0 then (1 + β)m > 1 + mβ.
(2) If −1 < β < 0 then (1 + β)m > 1 + mβ.
(3) If β = 0 then (1 + β)m = 1 + mβ.
The result follows from a combination of these three statements.]

(1) Suppose β > 0. Then, since β > 0 and 1 + β > 1, we have

(1 + β)m − 1 = β[(1 + β)m−1 + (1 + β)m−2 + · · · + (1 + β) + 1]

> β (1 + 1 + · · · + 1 + 1) = mβ
| {z }
m copies

Then (1 + β)m > 1 + mβ.

1
 (2) Suppose −1 < β < 0. Then, since −β > 0 and 0 < 1 + β < 1, we have

1 − (1 + β)m = −[(1 + β)m − 1] = (−β)[(1 + β)m−1 + (1 + β)m−2 + · · · + (1 + β) + 1]

< (−β) (1 + 1 + · · · + 1 + 1) = −mβ
| {z }
m copies

Then (1 + β)m > 1 + mβ.

(3) Suppose β = 0. Then (1 + β)m = 1 = 1 + mβ.

The result follows.

4. We are going to give some applications of Theorem (1), and various generalizations of Theorem (1). It deserves notice
that when combined with the Sandwich Rule (from your calculus course), Theorem (1) will yield some basic results
in mathematical analysis.
Sandwich Rule.
Let {un }∞ ∞ ∞
n=0 , {vn }n=0 , {wn }n=0 be infinite sequences in R.

Suppose that for any n ∈ N, un ≤ vn ≤ wn . Further suppose that {un }∞ ∞

n=0 , {wn }n=0 converge to the same limit, say,
ℓ in R.
Then {vn }∞
n=0 also converges to ℓ.

5. Theorem (2).
1
Let r be a non-zero real number. Suppose |r| < 1, and α = − 1. Then the statements below hold:
|r|

1
(a) α > 0, and |r|n < for any n ∈ N\{0, 1}.
nα
(b) lim rn = 0.
n→∞

X
n
1
(c) lim rk = .
n→∞ 1−r
k=0

Remark. Statement (c) immediately yields the result below on geometric progression (as stated in the handout
Arithmetic progression and geometric progression):

Let {bn }∞
n=0 be a geometric progression of non-zero real numbers, with common ratio r. Suppose |r| < 1.
b0
Then lim (b0 + b1 + b2 + · · · + bn ) = .
n→∞ 1−r

6. Proof of Theorem (2).

1
Let r be a non-zero real number. Suppose |r| < 1, and α = − 1.
|r|

1
(a) Note that > 1. Then α > 0.
|r|
Pick any n ∈ N\{0, 1}. By Bernoulli’s Inequality, we have (1 + α)n > 1 + nα.
1
Now note that |r| = .
1+α
1 1 1
Then |r|n = < < .
(1 + α)n 1 + nα nα
(b) [We apply the Sandwich Rule here.]
1 1
For any n ∈ N\{0, 1}, we have − < −|r|n ≤ rn ≤ |r|n < .
nα nα
1 1
Note that lim − = 0 and lim = 0.
n→∞ nα n→∞ nα

By the Sandwich Rule, lim rn = 0 exists and is equal to 0.

n→∞

2
 X
n
1 − rn+1 1 1
(c) For each n ∈ N, we have rk = = − · rn .
1−r 1−r 1−r
k=0
X
n
1
Then lim rk exists and is equal to .
n→∞ 1−r
k=0

7. Theorem (3).
The statements below hold:
(a) Let n ∈ N\{0, 1} and a ∈ R. Suppose a > 1. Then n(a − 1) < an − 1 < nan−1 (a − 1).
(b) Let n ∈ N\{0, 1} and b ∈ R. Suppose b > 1. Then bn − (b − 1)n < nbn−1 < (b + 1)n − bn .

8. Proof of Theorem (3). Construct the arguments as an exercise, using the roughwork below as a framework:

(a) i. For each a and n, we attempt to separately deduce the inequality an − 1 > n(a − 1) and the inequality
an − 1 < nan−1 (a − 1).
ii. An equivalent formulation of the former inequality is an > 1 + n(a − 1).
This suggests we apply Bernoulli’s Inequality appropriately.
 
1 1
iii. An equivalent formulation of the latter inequality is 1 − n < n 1 − .
a a
 
1 1
This can be further re-formulated as n > 1 + n −1 .
a a
This again suggests we apply Bernoulli’s Inequality appropriately.
(b) i. For each b and n, we attempt to separately deduce the inequality bn − (b − 1)n < nbn−1 and the inequality
(b + 1)n − bn > nbn−1 .
 n  n−1
b b 1
ii. An equivalent formulation of the former inequality is −1<n · .
b−1 b−1 b−1
This suggests we apply Statement (a) appropriately.
 n
b+1 1
iii. An equivalent formulation of the latter inequality is −1>n· .
b b
This again suggests we apply Statement (a) appropriately.

9. Theorem (4).

mp+1 X m
(m + 1)p+1 − 1
(a) Suppose p ∈ N\{0} and m ∈ N\{0, 1}. Then < kp < .
p+1 p+1
k=1

X
m  p
1 k 1
(b) Suppose p ∈ N\{0}. Then lim = .
m→∞ m m p+1
k=1

Proof of Theorem (4). Exercise. Apply Theorem (3) and the Sandwich Rule.
Remark. Statement (b) is at the core of the argument (from the definition of Riemann integrals) for the result
below:
Z x
xp+1
Suppose x ∈ R and p ∈ N\{0}. Then tp dt = .
0 p+1

10. Theorem (5).

The statements below hold:
cn − 1 cn−1 − 1
(a) Let n ∈ N\{0, 1} and c ∈ R. Suppose c > 1. Then > .
n n−1
dk − 1 dℓ − 1
(b) Let k, ℓ ∈ N\{0} and d ∈ R. Suppose d > 1, and k > ℓ. Then > .
k ℓ
Proof of Theorem (5). Exercise. (Apply Theorem (3) to prove Statement (a). Statement (b) is an immediate
consequence of Statement (a).)

11. Theorem (6).

The statements below hold:

3
 ar − 1
(a) Let r be a rational number, and a ∈ R. Suppose r > 1, and a > 1. Then > a − 1.
r
(b) Let r be a rational number, and β ∈ R. Suppose r > 1, and β > 0. Then (1 + β)r > 1 + rβ.

Proof of Theorem (6). Exercise. (Apply Theorem (6) to prove Statement (a). can you name some appropriate
positive integers k, ℓ and some real number d for which ar = dk and aℓ = d? Statement (b) is an immediate
consequence of Statement (a).)

12. Carefully modifying what has been done in Theorem (3), Theorem (5) and Theorem (6), we can obtain the result
below, which ‘generalizes’ Theorem (1) from ‘positive integral indices’ to ‘rational indices’.
Theorem (7). (Generalization of Bernoulli’s Inequality to ‘rational indices’.)
Let µ be a rational number, and β be a real number. Suppose µ ̸= 0 and µ ̸= 1, and β > −1. Then the statements
below hold:

(a) Suppose µ < 0 or µ > 1. Then (1 + β)µ ≥ 1 + µβ.

(b) Suppose 0 < µ < 1. Then (1 + β)µ ≤ 1 + µβ.
(c) In each of (a), (b), equality holds iff β = 0.

As a bonus we can also obtain the result below, which can be regarded as a generalization of Theorem (5):
Theorem (8).
Let s, t be rational numbers and c ∈ R. Suppose s > t > 1. The statements below hold:

cs − 1 ct − 1
(a) Suppose c > 1. Then > .
s t
1 − cs 1 − ct
(b) Suppose 0 < c < 1. Then > .
s t