Nested surds

A few months ago, I noted the following problem on a separate ‘blog for a textbook a colleague and I created to fill a need at our school.

Assuming the pattern continues forever, what is the value of \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}} ?

As you know, I encourage my students to be creative and solve problems in ways that make the most sense to them.  Following are three different approaches my students used when presented with this problem.  All three are essentially identical invocations of the problem’s self-similarity, but I liked the differences in how they approached the problem.

STOP!!!
SOLUTION ALERT!!
DO NOT READ ANY FURTHER IF YOU WANT TO SOLVE THESE PROBLEMS ON YOUR OWN.

Approach 1:
Notice that other than the outermost square root and addition, the entire problem is repeated inside the problem.  Therefore, if X=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}}, then

X=\sqrt{1+X}
X^2-X-1=0
X=\frac{1\pm\sqrt{5}}{2}

The value of X is obviously positive, so X=\frac{1+\sqrt{5}}{2}=\phi, the golden ratio!  How’s that for an unexpected surprise?!

Approach 2:
Two students thought in terms of sequences.  Each assumed a first term of a_1=\sqrt{1} and iterated using a_n=\sqrt{a_{n-1}}.

Recognizing the decimal equivalent of \phi, the first student concluded that the iterated surd appeared to be the golden ratio, but said in a class discussion the following day that while he had compelling evidence for an answer, he knew he hadn’t proven his solution.

Approach 3:
My second student who thought of a sequence approach reached the a_n=\sqrt{a_{n-1}} statement and realized that if she was able to iterate forever and this problem actually had an answer, then there would be a point where L=a_{n-1}=a_n where L is the limit of the iteration.  At this point, she wrote L=\sqrt{1+L} and proceeded as in Approach 1 to get L=\phi.

Some readers might argue that Approaches 1 and 3 really aren’t distinct, but I maintain that the students who used Approach 1 (the clear majority) were thinking purely algebraically and took advantage of the problem’s self-similarity, while the girl who used Approach 3 was thinking numerically and found her answer using limits.

Minor Extension:
For fun, I also assigned 1+\displaystyle\frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}.  At first, this problem looks very different from the one discussed above, but its closed form delighted some (and disturbed others!)  with its connection.  I hope you enjoy!

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