Nested radicals and its convergence

[flyingpig]

Homework Statement

This is supposed to be really easy, but I don't think my answer is good

Consider this

\sqrt{1 + \sqrt{1 + \sqrt{1 + ...}}}

I was hinted that a_{n + 1} = \sqrt{1 + a_n} for all n ≥ 0 and I am supposed to show that the sequence convergees

The Attempt at a Solution

Am I suppose to use a_{n +1} converges or a_n?

Since the nested radicals go on to infinity, wouldn't it be better to write it as

a_n = \sqrt{1 + a_n}

So that

a^2 _n = 1 + a_n

We get a quadratic and solve (on Maple) we get

\frac{1}{2}(\sqrt(5) + 1)

I rejected negative root because there is no way a negative root can occur in this sequence (we are just adding positive numbers and rooting it (I hope that's a word))

 

[HallsofIvy]

Homework Statement

This is supposed to be really easy, but I don't think my answer is good

Consider this

\sqrt{1 + \sqrt{1 + \sqrt{1 + ...}}}

I was hinted that a_{n + 1} = \sqrt{1 + a_n} for all n ≥ 0 and I am supposed to show that the sequence convergees

The Attempt at a Solution

Am I suppose to use a_{n +1} converges or a_n?

Those are just different numberings for the same sequence if one converge the other converges to the same thing.

Since the nested radicals go on to infinity, wouldn't it be better to write it as

a_n = \sqrt{1 + a_n}

No, that's not true for any finite value of n. What is true is that if \lim_{n\to\infty} a_n= a, then \lim_{n\to\infty}\sqrt{1+ a_n}= \sqrt{1+ \lim_{n\to\infty} a_n}= \sqrt{1+ a}.

So that

a^2 _n = 1 + a_n

Okay, except that that it should be the value of the limit, a, not a_n

We get a quadratic and solve (on Maple) we get

\frac{1}{2}(\sqrt(5) + 1)

I rejected negative root because there is no way a negative root can occur in this sequence (we are just adding positive numbers and rooting it (I hope that's a word))

Aw, c'mon! You use Maple to solve a quadratic equation? (Yes, that is the correct limit.)