Consider the following expression: $$x=\sqrt{2+{\sqrt{2+\sqrt{2+\ldots}}}}$$ It can easily be evaluated by realizing that: $$x=\sqrt{2+x}$$ After squaring both sides and solving the quadratic equation, we find that $x=2$. But my question is: out of curiosity, is there any other way to evaluate this nested radical? Perhaps one involving infinite series/products?
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1Of course your solution is incomplete. It shows only if it converges, then it converges to $2$. You also need to show somehow that it converges. – GEdgar Nov 21 '15 at 15:31
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And if we observe that $$ \sqrt{2}\approx 1.4142 $$ $$ \sqrt{2+\sqrt{2}}\approx 1.8478 $$ $$ \sqrt{2+\sqrt{2+\sqrt{2}}}\approx 1.9616 $$ $$ \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}\approx 1.9904 $$ $$ \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}\approx 1.9976? $$ – Nov 21 '15 at 15:33
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Hint Prove by induction that
$$\sqrt{2+{\sqrt{2+\sqrt{2+\ldots\sqrt{2}}}}}=2 \cos \left(\frac{\pi}{2^{n+1}}\right)$$
where $n$ is the number of roots.
P.S. If I am not mistaken, this can be interpreted geometrically as follows:
If $A_1A_2....A_{2^n}$ is a regular polynomial with $A_1A_2=1$ then $$A_1A_3=\sqrt{2+{\sqrt{2+\sqrt{2+\ldots\sqrt{2}}}}}$$
Now, as $n$ increases the angle $A_2$ gets closer and closer to $180$ and hence $A_1A_3$ gets closer and closer to $A_1A_2+A_2A_3$.
