Let {an} be the sequence defined recursively by a1=√(2) and an+1=√(2+an) for n ≥1. (a) List the

Let {an} be the sequence defined recursively by a1=√(2) and...

Let $\left\{a_{n}\right\}$ be the sequence defined recursively by $a_{1}=\sqrt{2}$ and $a_{n+1}=\sqrt{2+a_{n}}$ for $n \geq 1$. (a) List the first three terms of the sequence. (b) Show that $a_{n}<2$ for $n \geq 1$. (c) Show that $a_{n+1}^{2}-a_{n}^{2}=\left(2-a_{n}\right)\left(1+a_{n}\right)$ for $n \geq 1$. (d) Use the results in parts (b) and (c) to show that $\left\{a_{n}\right\}$ is a strictly increasing sequence. [Hint: If $x$ and $y$ are positive real numbers such that $x^{2}-y^{2}>0$, then it follows by factoring that $x-y>0$.] (c) Show that $\left\{a_{n}\right\}$ converges and find its limit $L$.

Let $\left\{a_{n}\right\}$ be the sequence defined recursively by $a_{1}=\sqrt{2}$ and $a_{n+1}=\sqrt{2+a_{n}}$ for $n \geq 1$.
(a) List the first three terms of the sequence.
(b) Show that $a_{n}<2$ for $n \geq 1$.
(c) Show that $a_{n+1}^{2}-a_{n}^{2}=\left(2-a_{n}\right)\left(1+a_{n}\right)$ for $n \geq 1$.
(d) Use the results in parts (b) and (c) to show that $\left\{a_{n}\right\}$ is a strictly increasing sequence. [Hint: If $x$ and $y$ are positive real numbers such that $x^{2}-y^{2}>0$, then it follows by factoring that $x-y>0$.]
(c) Show that $\left\{a_{n}\right\}$ converges and find its limit $L$.

Key Concepts

Convergence of Sequences

Convergence refers to the process wherein the terms of a sequence approach a specific limit as the index increases indefinitely. Understanding convergence is crucial in sequence analysis, especially when combined with properties like monotonicity and boundedness, which, by the monotone convergence theorem, ensure that a sequence converges to a finite limit.

Difference of Squares and Factoring

The difference of squares is an algebraic identity given by x^2 - y^2 = (x - y)(x + y). This factorization is useful in analyzing relationships between terms of a sequence, particularly when these differences reveal important information about the sequence’s behavior and monotonicity.

Monotonic Sequences

A monotonic sequence is one that is either entirely non-decreasing or non-increasing. Demonstrating the monotonicity of a sequence is vital because, when combined with boundedness, it guarantees convergence. In the context of recursively defined sequences, showing strict monotonicity often involves comparing successive terms.

Mathematical Induction

Mathematical induction is a proof method used to establish the truth of an infinite number of cases by proving a base case and then showing that if an arbitrary case holds, the next case must also hold. This technique is essential when proving properties, like bounds or monotonicity, that are asserted for every term in a recursively defined sequence.

Recursively Defined Sequences

These are sequences in which each term is defined in relation to one or more of its preceding terms, starting from an initial condition. This concept is critical in understanding how the sequence's behavior emerges step by step and in analyzing its long-term properties, such as convergence or boundedness.

Transcript

00:01 Given the sequence a n with a 1 is equal to square root of 2, a n can be defined inductively by 2 plus a, a n plus 1 is equal to square root of 2 plus a n, for any n is a positive natural number.

00:31 First, we are required to write down the first three terms of our sequence.

00:37 The first term has been given in this question, the second term by the definition will be equal to square root of 2 plus a 1, which will be square root of 2 plus square root of 2.

00:52 Third term will be equal to square root of 2 plus a 2, square root of 2 plus square root of 2 plus square root of 2.

01:03 So, a n can be represented as a very long square root thing.

01:14 We are required to show a n will be strictly less than 2 for any n.

01:19 Notice a 1 is equal to square root of 2, which is strictly less than 2.

01:24 We just want to prove it by induction.

01:28 And suppose for any k which is greater than 1 and less equal to n, we have a k is less than 2.

01:49 Now, let's consider k is equal to n plus 1.

01:53 Now, a n plus 1 by the definition will be equal to square root of 2 plus a n.

01:59 Now, by our inductive assumption, when k is equal to n, we know a n, a k will be strictly less than 2.

02:10 So, it will be strictly less than square root of 2 plus, right? which will be equal to square root of 4 and will be equal to 2.

02:27 Okay, so what do we prove? we make the inductive assumption and prove this is true for n plus 1 by our inductive hypothesis.

02:39 That means a n is strictly less than 2, must be held for any natural number.

02:54 This is by the induction.

02:58 So, we finish proof for the second part.

03:02 For the third part, we need to move on to show.

03:06 A n plus 1 square minus a n square is equal to 2 minus a n times 1 plus a n.

03:17 Okay, use the expression of a n plus 1.

03:22 We know a n plus 1 square will be just equal to 2 plus a n.

03:28 And the left side of our identity, which is equal to a n plus 1 square minus a n will be just equal to 2 plus a n minus a n.

03:45 And it's very easy for us to see as a function, any 2 plus x minus x1.

03:51 We can know this function has two solutions.

03:56 It has two roots.

03:57 The first one is 2, the second one is minus 1.

04:00 Use this fact, we can factorize it as 2 minus a n times 1 plus a n.

04:08 This can be factored.

04:09 As this equation has two distinct roots, we can separate it as 2 plus x times x plus 2 minus x.

04:25 Okay, then use the fact in c, we can show a n plus 1 is actually greater than a n.

04:39 Because all of the terms, all of our sequence, is all terms in our sequence are non -negative...

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