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

Community
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2 Answers2

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The answer given was completely wrong.

KENYER
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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.

Ahaan S. Rungta
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  • If I may please ask $-$ why did I get 2 upvotes on this on the same morning despite the 3 upvotes and the bounty? – Ahaan S. Rungta Oct 21 '14 at 18:54
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    Two downvotes, you mean? It might be because this thread was just linked to elsewhere. In short, your solution appears to be completely wrong; the quadratic formula yields $S=a$, which should imply (by your reasoning) that the entire expression is zero. But it's not; the difficulty here is in determining how quickly $S$ approaches zero, and comparing that with the multiplying factor. – Andrew Dudzik Oct 21 '14 at 22:52
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    This problem is hard, or at least, quite a bit harder than the approaches in this thread seem to imply. When $a=2$, the limit is actually $\pi$, and it is not clear how one can get there with things like the quadratic formula. For $a\neq 2$, it is very possible that there is a positive limit for this sequence, but without a closed form. – Andrew Dudzik Oct 21 '14 at 22:54
  • @Slade Yes, I meant 2 downvotes. Thanks for the explanation. 2 upvotes. :) – Ahaan S. Rungta Oct 21 '14 at 23:21