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The adjacent suites: Courses and corrected exercises

This article aims to present adjacent sequences through their definition, examples and corrected exercises. It is good to have basic knowledge about sequences, namely arithmetic sequences and geometric sequences.

Definition

Two suites (un) and Vn) are said to be adjacent if:

• The sequence (un) is increasing
• The sequence (vn) is decreasing
• The limit of their difference is zero:
\lim_{n \to +\infty} v_n - u_n = 0

Then we have the following theorem, called the theorem of adjacent sequences:

The sequences (un) and Vn) converge to the same limit.

Moreover, we can note the following property:

\forall n \in \mathbb{N}, u_0 \leq u_n \leq l \leq v_n \leq v_0

Examples

Consider the following two geometric sequences:

u_n = \dfrac{1}{2^n}, v_n =- \dfrac{1}{2^n}

We have :

• (un) is decreasing
• (vn) is increasing
• The limit of their difference is zero:
\lim_{n \to +\infty} u_n-v_n = 0

These two suites are therefore very adjacent.

Corrected exercises

Demonstration of the irrationality of e

The proof of the irrationality of e uses adjacent sequences

Exercise 39 (adjacent suites prep level)

Question 1

To show that these reals are well defined, it suffices to show that the elements are indeed positive.

We will show this existence by induction

• Initialization: a0 and B0 are well defined and positive
• Heredity: It is assumed that for a given n, an and Bn exist and are positive. So bn + 1 exists and is indeed positive as an arithmetic mean of positive terms. Moreover,
a_{n+1}= \sqrt{a_nb_n} \geq 0

And therefore does exist.
For the second part of the question, we will do it without recurrence. The case n = 0 is obvious. We will assume n > 0

As :

(\sqrt{a_{n-1}}-\sqrt{b_{n-1}})^2 \geq 0

We have :

a_{n-1} + b_{n-1} - 2 \sqrt{a_{n-1}b_{n-1}} \geq 0

So rearranging the terms, we get:

\dfrac{a_{n-1}+b_{n-1}}{2} \geq \sqrt{a_{n-1}b_{n-1}}

Which gives us:

b_n \geq a_n

We have shown in the first part of the question that

a_n \geq 0

Question 2

We have :

\begin{array}{ll} a_{n+1} - a_n& = \sqrt{a_nb_n}-a_n \\ &=\sqrt{a_n}(\sqrt{b_n}-\sqrt{a_n}) \\ &\ geq 0 \end{array}

So (an) is increasing

Urban artist

\begin{array}{ll} b_{n+1} - b_n& =\dfrac{b_n+a_n}{2}-b_n \\ &=\dfrac{a_n-b_n}{2} \\ &\leq 0 \ end{array}

So (bn) is decreasing. Moreover :

\begin{array}{ll} b_{n+1}-a_{n+1}& = \dfrac{a_n+b_n}{2} - \sqrt{a_nb_n}\\ & \leq \dfrac{a_n+b_n }{2} - \sqrt{a_na_n} \\ &=\dfrac{b_n-a_n}{2} \end{array}

We then have, by a recurrence left to the reader:

0 \leq b_n -a_n \leq \dfrac{ba}{2^n}

And so, by framing theorem:

\lim_{n \to +\infty} b_n-a_n = 0

The sequences (an) and Bn) are therefore well adjacent.
NB : The common limit of (an) and Bn) is called the arithmetic-geometric mean of a and b and it is denoted by M(a,b).

Complementary exercises

Here is a first exercise

Show that this pair of sequences are adjacent sequences

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