J. Let a0 and a1 be positive real numbers, and set an+2 =...

J. Let $a_0$ and $a_1$ be positive real numbers, and set $a_{n+2} = \sqrt{a_{n+1} + \sqrt{a_n}}$ for $n \ge 0$. (a) Show that there is $N$ such that for all $n \ge N$, $a_n \ge 1$. (b) Let $\epsilon_n = |a_n - 4|$. Show that $\epsilon_{n+2} \le (\epsilon_{n+1} + \epsilon_n)/3$ for $n \ge N$. (c) Prove that this sequence converges.

Transcript

00:01 Hello friends, for the first bit to prove that the sequence xn, n belongs to n is convergent we will show that it is both bounded and monotonic.

00:20 First prove the boundedness given that xn is less than equal to m for any n belongs to n.

00:32 We can conclude that the sequence is bounded above by m.

00:36 Additionally, since xn is a sum of an absolute value, it cannot be negative.

00:41 So it is bounded below by 0.

00:49 Therefore, the sequence is bounded.

00:56 That is, xn lies between 0 to m for all n belongs to n.

01:04 Now monotonicity.

01:05 Let's consider two arbitrary indices m and n, where m is less than n.

01:14 We need to show that xm is less than equal to xn.

01:18 Using the definition of xn, we have xn equal to summation i equal to 1 to n, xi plus 1 minus xi.

01:28 Now let's split the sum into two parts.

01:30 That is, summation i equal to 1 to n, xi plus 1 minus xi plus summation i equal to m to m plus 1 to n, that is, xi plus 1 minus xi.

01:54 The first sum represents xm and the second sum represents the additional terms from m plus 1 to n.

02:02 Since the absolute value function is non -negative, the sum of additional terms can only increase the value of xn.

02:09 Thus, we have xn is greater than equal to xm, which implies that xn n belongs to n is non -decreasing.

02:18 Since xn n belongs to n is bounded and non -decreasing, so it is constant, convergent.

02:41 So it is convergent.

02:42 Now moving forward to the b bit to prove that the sequence xn n belongs to n is convergent using cauchy criterion, we need to show that it satisfies cauchy property.

02:58 The cauchy criterion states that a sequence an n belongs to n is convergent if and only for every epsilon is greater than equal to zero, there exist an capital n belongs to natural number such that for all m n is greater than equal to n and an minus m mod is less than equal to epsilon.

03:19 Let's consider two arbitrary indices m and n such that m is less than equal to n...

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