Verified answer

We are given

a_(n+1) = 1 + 1 / a_n.

Since a_n → L as n → ∞,we can take lim n → ∞ of both sides of the equation above to get

L = 1 + 1 / L.

L^2 - L - 1 = 0

So L = (1±√5) / 2, except L>0. Therefore L = (1+√5) / 2.

Start with trying to cook up some examples. See what they tell you.

a0 = 1

a1 = 1 + 1 = 2

a2 = 1 + 1/2 = 3/2

a3 = 1 + 2/3 = 5/3

a4 = 1 + 3/5 = 8/5

a5 = 1 + 5/8 = 13/8

Huh. That's kind of interesting already. I happen to recognize those numbers as Fibonacci numbers, and I happen to know the limit of their ratio is the Golden Ratio. Let's try a general starting point and see if the Fibonacci numbers fall out. Or something.

a2 = 1 + 1/a1 = (a1 + 1)/a1

a3 = 1 + a1/(1 + a1) = (1 + a1 + a1)/(1 + a1) = (1 + 2a1)/(1 + a1)

a4 = 1 + (1 + a1)/(1 + 2a1) = (1 + 2a1 + 1 + a1)/(1 + 2a1) = (2 + 3a1)/(1 + 2a1)

a4 = 1 + (1 + 2a1)/(2 + 3a1) = (2 + 3a1 + 1 + 2a1)/(2 + 3a1) = (3 + 5a1)/(2 + 3a1)

And all those coefficients are actually Fibonacci numbers, but I'm not sure offhand what to say about this general sequence.

The relationship between these numbers and the Fibonacci sequence you'd have to prove by showing the relationship of your recurrence relation to the Fibonacci one: a_(n+1) = a_(n-1) + a_n

Something like this: a_(n+1) = 1 + 1/a_n = (a_n + 1)/a_n and then substituting a_n = (a_{n-1} + 1)/a_{n-1} , a_n + 1 = (a_{n-1} + 1 + a_{n-1})/a_{n-1}

I'm just thinking out loud here but I think this is the right general direction.

Edit: Hah! And while I was doing that, somebody else came up with an elegant argument that goes straight to the Golden Ratio.