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If $a$, $b$ are positive quantities such that ($a < b$) and if $$a_1=\frac{a+b}{2},b_1=\sqrt{a_1b},\ a_2=\frac{a_1+b_1}{2},b_2=\sqrt{a_2b_1},\ldots,\\ a_n=\frac{a_{n-1}+b_{n-1}}{2},b_n=\sqrt{a_{n}b_{n-1}}$$

Prove that $$\lim_{n \to \infty}b_n=\dfrac{\sqrt{b^2-a^2}}{\cos^{-1}(b/a)}$$

I tried to show that as $n\to \infty,a_n=b_n\implies a_{n-1}=b_{n-1}$

But I do not find this result quite promising. Please help. Some hint/solution is appreciated.

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Saradamani
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  • Do you mean $\arccos(a/b)$? – Did May 09 '18 at 09:36
  • @Did yes I mean that exactly. – Saradamani May 09 '18 at 09:38
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    Then correct your post? And while we are at it, could you add some personal tries that would be more convincing than the unique present, rather bizarre, one? – Did May 09 '18 at 09:43
  • @Did But that is exactly what is written in the question paper of the test book I follow. $cos^{-1}\dfrac{b}{a}$. If you want I can post its scanned picture. My trials did not look promising to me-pages of efforts but ultimately I could not get anywhere near. Thats why I thought it be best to leave that for the answerer to decide. – Saradamani May 09 '18 at 09:46
  • A typo in the test book then? 'Cause what is $\arccos x$ when $x>1$, one wonders... – Did May 09 '18 at 11:03

1 Answers1

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Hint

You have $$\begin{cases} a_n=\frac{a_{n-1}+b_{n-1}}{2}\\ b_n=\sqrt{a_nb_{n-1}} \end{cases}$$

The second equation gives $a_n=\frac{b_n^2}{b_{n-1}}$. Replace this quantity in the first one, set $z_n=\frac{b_n}{b_{n-1}}$ and see what happen.

Surb
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