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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 $x_n$

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 $(x_{n+1}) - (x_n) \gt 0$?

Trying2LearnMath
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    Welcome to Math.SE! I think you misunderstood a key aspect of this site. This is more about helping you find the right answer then giving you answers to arbitrary problems. For example, if you were to give us what you've tried, where you've looked, what you've read to answer the problem, and describe why you aren't satisfied/doubt what you've come across, I'm sure others would help. Also, choosing a descriptive title that described the content would help. For more, I recommend reading How to ask a good question? – davidlowryduda Mar 02 '15 at 00:28
  • @mixedmath May I borrow this for future use? – Zubin Mukerjee Mar 02 '15 at 00:49
  • @ZubinMukerjee Go for it! – davidlowryduda Mar 02 '15 at 02:06
  • Part 3 should begin with, “Let $\varepsilon > 0$.” – chharvey Jun 30 '15 at 04:53
  • related: http://math.stackexchange.com/questions/115501/sqrtc-sqrtc-sqrtc-cdots-or-the-limit-of-the-sequence-x-n1-sq – chharvey Jun 30 '15 at 17:29

3 Answers3

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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

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

RRL
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  • It’s bounded above and increasing… what’s the limit? – chharvey Jun 30 '15 at 04:48
  • @chharvey: Since the sequence converges, we can take the limit of both sides of the recursion. All subsequences have the same limit and the square root function is continuous. Hence $L = \lim x_{n+1} = \lim \sqrt{6+x_n} = \sqrt{6 + L}$. This implies $L^2 - L - 6 =0$. The sequence starting with $x_1 = \sqrt{6}$ increases and converges to the positive root $L = 3$. – RRL Jun 30 '15 at 06:21
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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 $x_n = x_{n+1}$.

DanielV
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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 $({1/2})(1+\sqrt{1+4*6})=3.$

NoChance
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