Home » Bernoulli's Inequality

# Determine whether ∑ 1 / n1 + 1/n converges

Test the following series for convergence:

The series diverges.

Proof. First, we write

Then, we know for all . (We can deduce this from the Bernoulli inequality, with . We proved the Bernoulli inequality in this exercise, Section I.4.10, Exercise #14.) Therefore, and we have

Since the series diverges we have established the divergence of the given series

# Prove properties of the Bernoulli polynomials

The Bernoulli polynomials are defined by

1. Find explicit formulas for the first Bernoulli polynomials in the cases .
2. Use mathematical induction to prove that is a degree polynomial in , where the degree term is .
3. For prove that .
4. For prove that

5. Prove that

for .

6. Prove that for ,

7. Prove that for ,

Now, using the integral condition to find ,

Thus,

Next, using this expression for we have

Using the integral condition to find ,

Thus,

Next, using this expression for we have

Using the integral condition to find ,

Thus,

Next, using this expression for we have

Using the integral condition to find ,

Thus,

Finally, using this expression for we have

Using the integral condition to find ,

Thus,

2. Proof. We have shown in part (a) that this statement is true for . Assume then that the statement is true for some positive integer , i.e.,

Then, by the definition of the Bernoulli polynomials we have,

where for . Then, taking the integral of this expression

Hence, the statement is true for the case ; hence, for all positive integers

3. Proof. From the integral property in the definition of the Bernoulli polynomials we know for ,

Then, using the first part of the definition we have ; therefore,

Thus, we indeed have

4. Proof. The proof is by induction. For the case we have

Therefore,

Since , the stated difference equation holds for . Assume then that the statement holds for some positive integer . Then by the fundamental theorem of calculus, we have

Therefore,

Hence, the statement is true for the case , and so it is true for all positive integers

5. Proof. (Let’s assume Apostol means for to be some positive integer.) First, we use the definition of the Bernoulli polynomials to compute the integral,

Now, we want to express the numerator as a telescoping sum and use part (d),

Thus, we indeed have

6. Proof.

Incomplete. I’ll try to fix parts (f) and (g) soon(ish).

# Prove a generalization of Bernoulli’s inequality

For real numbers with for all and all of the having the same sign, prove

As a special case let and prove Bernoulli’s inequality,

Finally, show that if then equality holds only when .

Proof. The proof is by induction. For the case , we have,

so the inequality holds for .
Assume then that the inequality holds for some . Then,

But, since every must have the same sign (thus, and must have the same sign, so the product is positive). Thus,

Hence, the inequality holds for the case ; and therefore, for all

Now, if where and we apply the theorem above to obtain Bernoulli’s inequality,

Claim: Equality holds in Bernoulli’s inequality if and only if .
Proof.
If then , so indeed equality holds for . Next, we use induction to show that if , then the inequality must be strict. (Hence, equality holds if and only if .)
For the case , on the left we have,

since for . So, the inequality is strict for the case . Assume then that the inequality is strict for some . Then,

Where the final line follows since and implies . Therefore, the inequality is strict for all if .

Hence, the equality holds if and only if