Find the value of : $\lim_{n\to\infty}(2a)^{\frac{n}{2}}\sqrt{a-\sqrt{a(a-1)+\sqrt{a(a-1)+\cdots}}}$

Question

find the limit $$\lim_{n\to\infty}(2a)^{\frac{n}{2}}\underbrace{\sqrt{a-\sqrt{a(a-1)+\sqrt{a(a-1)+\cdots}}}}_{n \textrm{ square roots}}$$

My try: I know this Find $\lim_{n\to\infty}2^n\underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\dots+\sqrt2}}}}_{n \textrm{ square roots}}$.

and this problem just

$$\underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\dots+\sqrt2}}}}_{n \textrm{ square roots}}=4\left|\sin{\frac{\theta}{2^n}}\right|$$ so follow this limit is easy to find it.

But $$\underbrace{\sqrt{a-\sqrt{a(a-1)+\sqrt{a(a-1)+\cdots}}}}_{n \textrm{ square roots}}=?$$

Answer 1

The answer given was completely wrong.

Answer 2

Solution Outline (I will leave the details for you to fill).

So, we consider this: $$ (2a)^{\frac{n}{2}}\underbrace{\sqrt{a-\sqrt{a(a-1)+\sqrt{a(a-1)+\cdots}}}}_{n \textrm{ square roots}}, $$ where $n \to \infty$. We assert that the number of square roots that are nested is indeed just infinity. So, we consider this: $$ S = \sqrt{a(a-1)+\sqrt{a(a-1)+\cdots}}, $$ where, here, there are an infinite number of nested square roots. Then, $$S^2=a(a-1)+S.$$Then, you can solve for $S$ using the quadratic formula. our original expression is, remember the following: $$ (2a)^{\frac{n}{2}}\sqrt{a-S}. $$You can get $S$ using the quadratic formula and substitute into this expression. Then, finding the limit of the resulting expression will not be trivial but it won't be impossible either.