Help!

This computer summation for about 25 terms shows that x converges to sqrt(2):

c=1E100; listforeach(b,reverse(1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2), c=b + 1/c)

x =1.4142135623730950488016887242097

You  could also solve it algebraically in this way:

Solve for x:
1/(1 + x) + 1 = x

Write the left hand side as a single fraction.
Bring 1/(x + 1) + 1 together using the common denominator x + 1:
(x + 2)/(x + 1) = x

Multiply both sides by a polynomial to clear fractions.
Multiply both sides by x + 1:
x + 2 = x (x + 1)

Write the quadratic polynomial on the right hand side in standard form.
Expand out terms of the right hand side:
x + 2 = x^2 + x

Move everything to the left hand side.
Subtract x^2 + x from both sides:
2 - x^2 = 0

Isolate terms with x to the left hand side.
Subtract 2 from both sides:
-x^2 = -2

Multiply both sides by a constant to simplify the equation.
Multiply both sides by -1:
x^2 = 2

Eliminate the exponent on the left hand side.
Take the square root of both sides:
x = sqrt(2) 

Thanks for that great answer guest, 

I will just try to show a little clearer what I think you have done.

First add 1 to both sides.

\(x+1 = 2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}\)

now

\(x+1 = 2 + \cfrac{1}{x+1}\\ x-1 = \cfrac{1}{x+1}\\ x^2-1 = 1\\ x^2=2\\ x=\pm\sqrt2 \\ \text{But everything is positive on the RHS so } x>-1\\ x=\sqrt2\)