How can I show that the continued fraction N= 1+(1/(2+1/2+1/2+1/2+...) converges.?

Question

The answer is (2)^(1/2) but I do not know how to explain the fact that I assumed it converge in order to solve for it.

Answer 1

Notice that you can rewrite this as:
N = 1 + 1 / (2 + 1/(2+ 1/2+... )
N - 1 = 1 / (2 + 1/(2+ 1/2+... )

Since what's on the right side also repeats within itself, you can say:
N - 1 = 1 / (2 + [N - 1])

Now solve that for N:
(N - 1)(2 + [N - 1]) = 1
(N - 1)(N + 1) = 1
N^2 - 1 = 1
N^2 = 2
N = √2

Answer 2

Let
m = N - 1

Solve for m and add 1.

m = 1/(2 + m)

2m + m² = 1
m² + 2m - 1 = 0

m = {-2 ± √[2² - 4*1*(-1)]} / 2
m = {-2 ± √(4 + 4)} / 2
m = {-2 ± √8} / 2
m = {-2 ± 2√2} / 2
m = -1 ± √2

Since the number is positive, the negative solution is rejected.

m = -1 + √2

N = m + 1 = (-1 + √2) + 1 = √2

Answer 3

i've got faith the respond starts off like this: ? nx^n = x * ?d/dx (x^n) = x * d/dx (?x^n) via fact |x|