Calculating $\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots +\sqrt{2}}}}}_{100\;\text{times}}$ without a calculator?

Question

Is there any way to find the summation up to $100$ times without using calculator?

$$\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots +\sqrt{2}}}}}_{100\;\text{times}}$$

I know that if the above summation would be up to $\infty$ then we can find the summation up to infinity very easily.

First we will take the summation to be $x$ and then we will get a quadratic like $x^2-x-2=0$. Now we have to solve the quadratic and get the value of the summation very easily and the value of the summation will be $2$.

But how to find the summation in this case for 100 times 2?

Kindly help me out with this question.

Answer 1

Hint: Let $f(x)=2\cos x$. From the half-angle formula for cosine and all terms being positive, render $f(x/2)=\sqrt{2+f(x)}$.

Can you continue from there?

To the significant figures shown: $1.9999999999999999999999999999999999999999999999999999999999985$.

Answer 2

Look at the sequane $$\cos \left(\frac{\pi }{2^2}\right)=\frac 12 \sqrt 2$$ $$\cos \left(\frac{\pi }{2^3}\right)=\frac 12 \sqrt{2+\sqrt{2}}$$ $$\cos \left(\frac{\pi }{2^4}\right)=\frac 12 \sqrt{2+\sqrt{2+\sqrt{2}}}$$ $$\cos \left(\frac{\pi }{2^5}\right)=\frac 12 \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}$$

which are obtained using systemetically the double angle formula for the cosine.

So

$$\cos\left(\dfrac{\pi}{2^n}\right)== \frac 12 \underbrace{\sqrt{2+\sqrt{2+ \sqrt{2+\cdots \sqrt{2}}}}}_n$$ Therefore $$\underbrace{\sqrt{2+\sqrt{2+ \sqrt{2+\cdots \sqrt{2}}}}}_n=2 \cos\left(\dfrac{\pi}{2^n}\right)$$

So, using the series expansion of the cosine, to second order $$2 \cos\left(\dfrac{\pi}{2^n}\right)=2 \left( 1-\frac {\pi^2}{2^{ 2 n+1}}+\frac 13 \frac {\pi^4}{2^{4n+3}} \right)$$

$$\underbrace{\sqrt{2+\sqrt{2+ \sqrt{2+\cdots \sqrt{2}}}}}_n\sim 2-\frac {\pi^2}{4^n}$$

For $n=100$, you should obtain $119$ exact significant figures.