Using induction to study the sequence $\sqrt{6} , \sqrt{6 +\sqrt{6}}, \dots$

Question

For the given sequence

$\sqrt{6} , \sqrt{6 +\sqrt{6}},\sqrt{6+\sqrt{6+\sqrt{6}}} $ ...

1. Use induction to show the sequence is bounded above by 3
2. Use induction to show $x_n $ is increasing
3. Find the limit of $x_n$

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Well I know for a fact I am overthinking the induction steps.

For part two, I know I need to show that $x_{n + 1} \gt x_n$ for all n, correct? Would this be as simple as showing $(x_{n+1}) - (x_n) \gt 0$?

Answer 1

I suggest formulating part 2 as $\sqrt{6 + x} > x$, but first you must (inductively) establish that $0 < x < 3$.

For part 2, first you have to establish that a limit exists. That was what parts 1 and 2 were for: increasing , but having an upper limit, so a limit exists.

To find the limit, solve $x_n = x_{n+1}$.

Answer 2

This sequence has $x_1 = \sqrt{6}$ and satisfies the recursion

$$x_{n+1} = \sqrt{6 + x_n}.$$

Hence $x_1 < 3$ and

$$x_n < 3 \implies x_{n+1} = \sqrt{6 + x_n} < \sqrt{6+3}= 3.$$

Use a similar inductive argument to prove the sequence is increasing.

We have $x_2 > x_1$ and

$$x_n > x_{n-1} \implies \sqrt{6 + x_n} > \sqrt{6 + x_{n-1}} \implies x_{n+1} > x_n$$

Answer 3

On how to get the limit (I assume you want to get the value of the generalized form) of $\sqrt{6+\sqrt{6+\sqrt{6+...}}}$:

You may want to use the concept of Infinitely nested radicals.

The answer would be $({1/2})(1+\sqrt{1+4*6})=3.$