#### Question

Show that \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots \infty } } } } = 2 when the layers of square root go to infinite

Collected in the board: Square Root

Steven Zheng posted 4 days ago

Let

x = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots \infty } } } }

when the layers of square root go to infinite

which makes no difference starting to count from the second square root.

Then, we get the following equation

x = \sqrt{2+x}

Then,

x > 0

Square both sides results in a quadratic equation

x^2-x-2=0

Factor the left hand side

(x-2)(x+1) = 0

Then,

x = 2

or

x = -1 \:(\text{cancel})

In summary,

\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots \infty } } } } = 2

Steven Zheng posted 4 days ago

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