Tag: Bernoulli’s Inequality

Determine whether ∑ 1 / n^{1 + 1/n} converges

by RoRi

March 29, 2016

Test the following series for convergence:

$\sum_{n=1}^{\infty} \frac{1}{n^{1 + \frac{1}{n}}}.$

The series diverges.

Proof. First, we write

$\frac{1}{n^{1+\frac{1}{n}}} = \frac{1}{n \cdot n^{\frac{1}{n}}}.$

Then, we know $n < 2^n$ for all $n \geq 1$ . (We can deduce this from the Bernoulli inequality, $(1+x)^n \geq 1 + nx$ with $x = 1$ . We proved the Bernoulli inequality in this exercise, Section I.4.10, Exercise #14.) Therefore, $n^{\frac{1}{n}} < 2$ and we have

$\frac{1}{n \cdot n^{\frac{1}{n}}} > \frac{1}{2n}.$

Since the series $\sum \frac{1}{2n}$ diverges we have established the divergence of the given series

$\sum_{n=1}^{\infty} \frac{1}{n^{1+\frac{1}{n}}}. \qquad \blacksquare$

Prove properties of the Bernoulli polynomials

by RoRi

October 31, 2015

The Bernoulli polynomials are defined by

$P_0(x) = 1; \qquad P'_n(x) = n P_{n-1} (x) \qquad \text{and} \qquad \int_0^1 P_n (x) \, dx = 0 \qquad \text{if } n \geq 1.$

Find explicit formulas for the first Bernoulli polynomials in the cases $n = 1,2,3,4,5$ .
Use mathematical induction to prove that $P_n(x)$ is a degree $n$ polynomial in $x$ , where the degree $n$ term is $x^n$ .
For $n \geq 2$ prove that $P_n (0) = P_n (1)$ .
For $n \geq 1$ prove that
$P_n (x+1) - P_n (x) = nx^{n-1}.$
Prove that
$\sum_{r=1}^{k-1} r^n = \int_0^k P_n (x) \, dx = \frac{P_{n+1}(k) - P_{n+1}(0)}{n+1}$

for $n \geq 2$ .
Prove that for $n \geq 1$ ,
$P_n (1-x) = (-1)^n P_n (x).$
Prove that for $n \geq 1$ ,
$P_{2n+1} (0) = 0 \qquad \text{and} \qquad P_{2n-1}\left( \frac{1}{2} \right) = 0.$

We start with the initial condition $P_0(x) = 1$ . This gives us
$P'_1 (x) = 1 \cdot P_0(x) = 1 \quad \implies \quad P_1 (x) = \int dx = x + C_1.$

Now, using the integral condition to find $C_1$ ,

$\int_0^1 (x+C_1) \, dx = 0 \quad \implies \quad \left( \frac{1}{2}x^2 + C_1 x \right) \Bigr \rvert_0^1 = 0 \quad \implies \quad C_1 &= -\frac{1}{2}.$

Thus,

$P_1 (x) = x -\frac{1}{2}.$

Next, using this expression for $P_1(x)$ we have

$P'_2 (x) = 2 \cdot \left( x- \frac{1}{2} \right) = 2x - 1 \quad \implies \quad P_2 (x) = x^2 - x + C_2.$

Using the integral condition to find $C_2$ ,

$\int_0^1 (x^2 - x + C_2) \, dx = 0 \quad \implies \quad \frac{1}{3} - \frac{1}{2} + C_2 = 0 \quad \implies \quad C_2 = \frac{1}{6}.$

Thus,

$P_2 (x) = x^2 - x + \frac{1}{6}.$

Next, using this expression for $P_2 (x)$ we have

$P'_3 (x) = 3 \cdot \left( x^2 - x + \frac{1}{6} \right) = 3x^2 - 3x + \frac{1}{2} \quad \implies \quad P_3(x) = x^3 - \frac{3}{2}x^2 + \frac{1}{2}x + C_3.$

Using the integral condition to find $C_3$ ,

$\int_0^1 \left( x^3 - \frac{3}{2}x^2 + \frac{1}{2}x + C_3\right) \, dx = 0 \quad \implies \quad \frac{1}{4} - \frac{1}{2} + \frac{1}{4} + C_3 = 0 \quad \implies \quad C_3 = 0.$

Thus,

$P_3 (x) = x^3 - \frac{3}{2}x^2 + \frac{1}{2} x.$

Next, using this expression for $P_3 (x)$ we have

$P'_4(x) = 4 \cdot \left( x^3 - \frac{3}{2}x^2 + \frac{1}{2}x \right) = 4x^3 - 6x^2 + 2x \quad \implies \quad P_4 (x) = x^4 - 2x^3 + x^2 + C_4.$

Using the integral condition to find $C_4$ ,

$\int_0^1 \left( x^4 - 2x^3 + x^2 + C_4 \right) \, dx = 0 \quad \implies \quad \frac{1}{5} - \frac{1}{2} + \frac{1}{3} + C_4 = 0 \quad \implies \quad C_4 = \frac{1}{30}.$

Thus,

$P_4 (x) = x^4 - 2x^3 + x^2 - \frac{1}{30}.$

Finally, using this expression for $P_4 (x)$ we have

$P'_5(x) = 5 \cdot \left( x^4 - 2x^3 + x^2 - \frac{1}{30} \right) = 5x^4 - 10x^3 + 5x^2 - \frac{1}{6} \quad \implies \quad P_5 (x) = x^5 - \frac{5}{2} x^4 + \frac{5}{3}x^3 - \frac{1}{6}x + C_5.$

Using the integral condition to find $C_5$ ,

$\int_0^1 \left( x^5 - \frac{5}{2} x^4 + \frac{5}[3} x^3 - \frac{1}{6} x + C_5 \right) \, dx = \frac{1}{6} - \frac{1}{2} + \frac{5}{12} - \frac{1}{12} + C_5 = 0 \quad \implies \quad C_5 = 0.$

Thus,

$P_5 (x) = x^5 - \frac{5}{2}x^4 + \frac{5}{3}x^3 - \frac{1}{6}x.$
Proof. We have shown in part (a) that this statement is true for $n = 0, 1, \ldots, 5$ . Assume then that the statement is true for some positive integer $m$ , i.e.,
$P_m (x) = x^m + \sum_{k=0}^{m-1} c_k x^k.$

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

$P_{m+1}' (x) = (m+1) \cdot \left( x^m + \sum_{k=0}^{m-1} a_k x^k \right) = (m+1)x^m + \sum_{k=0}^{m-1} b_k x^k,$

where $b_k = (m+1)a_k$ for $k = 1, \ldots, m-1$ . Then, taking the integral of this expression

$P_{m+1} (x) = \int \left( (m+1)x^m + \sum_{k=0}^{m-1} b_k x^k \right) \, dx = x^{m+1} + \sum_{k=0}^m \frac{b_k}{k+1} x^{k+1}.$

Hence, the statement is true for the case $m+1$ ; hence, for all positive integers $n. \qquad \blacksquare$
Proof. From the integral property in the definition of the Bernoulli polynomials we know for $n \geq 1$ ,
$\int_0^1 P_n (x) \, dx = 0 \quad \implies \quad \int_0^1 (n+1) P_n (x) \, dx = 0.$

Then, using the first part of the definition we have $P'_{n+1} (x) = (n+1) P_n (x)$ ; therefore,

$0 = \int_0^1 (n+1) P_n (x) \, dx = \int_0^1 P'_{n+1} (x) \, dx = P_{n+1}(1) - P_{n+1}(0).$

Thus, we indeed have

$P_{n+1} (1) = P_{n+1}(0). \qquad \blacksquare$
Proof. The proof is by induction. For the case $n = 1$ we have
$P_1 (x) = x - \frac{1}{2}, \qquad \text{and} \qquad P_1 (x+1) = x + \frac{1}{2}.$

Therefore,

$P_1 (x+1) - P_1 (x) = 1.$

Since $nx^{n-1} = 1 \cdot x^0 = 1$ , the stated difference equation holds for $n =1$ . Assume then that the statement holds for some positive integer $m$ . Then by the fundamental theorem of calculus, we have

$P_{m+1} (x) = \int_0^x P'_{m+1}(t) \, dt = (m+1) \int_0^x P_m (t) \, dt.$

Therefore,

$\begin{align*} P_{m+1}(x+1) - P_{m+1}(x) &= (m+1) \left( \int_0^{x+1} P_m (t) \, dt - \int_0^x P_m (t) \, dt \right) \\[9pt] &= (m+1) \left( \int_0^1 P_m (t) \, dt + \int_1^{x+1} P_m (t) \, dt - \int_0^x P_m (t) \, dt \right) \\[9pt] &= (m+1) \left( 0 + \int_0^x P_m (t+1) \, dt - \int_0^x P_m (t) \, dt \right) &( \text{Integral condition})\\[9pt] &= (m+1) \left( \int_0^x (P_m (t+1) - P_m (t)) \, dt\right) \\[9pt] &= (m+1) \left( \int_0^x mt^{m-1} \, dt \right)&(\text{Induction Hypothesis})\\[9pt] &= (m+1) x^m. \end{align*}$

Hence, the statement is true for the case $m+1$ , and so it is true for all positive integers $n. \qquad \blacksquare$
Proof. (Let’s assume Apostol means for $k$ to be some positive integer.) First, we use the definition of the Bernoulli polynomials to compute the integral,
$\begin{align*} \int_0^k P_n (x) \, dx &= \int_0^k \frac{1}{n+1} P'_{n+1} (x) \, dx \\[9pt] &= \frac{P_{n+1}(k) - P_{n+1}(0)}{n+1}. \end{align*}$

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

$\begin{align*} P_{n+1}(k) - P_{n+1}(0) &= \sum_{r = 1}^{k-1} \left( P_{n+1}(r + 1) - P_{n+1}(r) \right) \\[9pt] &= \sum_{r=0}^{k-1} \left( (n+1)r^n \right) \\ &= (n+1) \sum_{r=1}^{k-1} r^n. \end{align*}$

Thus, we indeed have

$\sum_{r=1}^{k-1} r^n = \int_0^k P_n (x) \, dx = \frac{P_{n+1}(k) - P_{n+1}(0)}{n+1}. \qquad \blacksquare$
Proof.

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

Prove a generalization of Bernoulli’s inequality

by RoRi

July 10, 2015

For real numbers $a_1, \ldots, a_n$ with $a_i > -1$ for all $i = 1, \ldots, n$ and all of the $a_i$ having the same sign, prove

$(1+a_1)(1+a_2) \cdots (1+a_n) \geq 1 + a_1 + a_2 + \cdots + a_n.$

As a special case let $a_1 = \cdots = a_n = x$ and prove Bernoulli’s inequality,

$(1+x)^n \geq 1 + nx.$

Finally, show that if $n > 1$ then equality holds only when $x = 0$ .

Proof. The proof is by induction. For the case $n =1$ , we have,

$(1+a_1) = 1 + a_1 \qquad \implies \qquad 1+a_1 \geq 1+a_1,$

so the inequality holds for $n=1$ .
Assume then that the inequality holds for some $n = k \in \mathbb{Z}_{>0}$ . Then,

$\begin{align*} &&(1+a_1) \cdots (1+a_k) &\geq 1 + a_1 + \cdots + a_k \\ \implies&& (1+a_1) \cdots (1+a_k)(1+a_{k+1}) &\geq (1+a_1 + \cdots + a_k)(1+a_{k+1})\\ &&&\geq (1+a_1+ \cdots + a_{k+1})+a_{k+1}(a_1 + \cdots + a_k). \end{align*}$

But, $a_{k+1}(a_1 + \cdots + a_k) \geq 0$ since every $a_i$ must have the same sign (thus, $a_{k+1}$ and $(a_1 + \cdots + a_k)$ must have the same sign, so the product is positive). Thus,

$(1+a_1) \cdots (1+a_{k+1}) \geq 1+a_1 + \cdots + a_{k+1}.$

Hence, the inequality holds for the case $k+1$ ; and therefore, for all $n \in \mathbb{Z}_{>0}. \qquad \blacksquare$

Now, if $a_1 = \cdots = a_n = x$ where $x > -1$ and $x \neq 0$ we apply the theorem above to obtain Bernoulli’s inequality,

$(1+a_1) \cdots (1+a_n) = (1+x)^n \geq 1+a_1 + \cdots + a_n = 1+n \cdot x.$

Claim: Equality holds in Bernoulli’s inequality if and only if $x = 0$ .
Proof.
If $x = 0$ then $(1+x)^n = 1 = 1 + n \cdot x$ , so indeed equality holds for $x = 0$ . Next, we use induction to show that if $x \neq 0$ , then the inequality must be strict. (Hence, equality holds if and only if $x = 0$ .)
For the case $n=2$ , on the left we have,

$(1+x)^2 = 1+2x + x^2 > 1+2x$

since $x^2 > 0$ for $x \neq 0$ . So, the inequality is strict for the case $n =2$ . Assume then that the inequality is strict for some $n = k \in \mathbb{Z}_{>1}$ . Then,

$\begin{align*} (1+x)^k > 1+kx\ &\implies & (1+x)^k (1+x) &> (1+kx)(1+x) &(x>-1 \implies 1+x>0)\\ &\implies & (1+x)^{k+1} &> 1 + (k+1)x + kx^2 \\ &\implies & (1+x)^{k+1} &> 1+(k+1)x. \end{align*}$