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Let $a_1=2$, $a_2=8$, $a_n=4a_{n-1}-a_{n-2}$, $n=3,4,5,\ldots$. Prove: $$\sum_{n=1}^\infty \mathrm{arccot} (a_n^2)=\frac{\pi}{12} $$

My attempt: I have worked out $a_n=\frac{\left(2+\sqrt{3}\right)^n-\left(2-\sqrt{3}\right)^n}{\sqrt{3}}$, but I do not know how to do it afterwards. Can anyone give me some suggestions? Thank you.

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I haven't tried, but there are formulas for $\mathrm{arccot}(a) + \mathrm{arccot}(b)$. You could try to work out by induction what the partial sums are. –  Najib Idrissi Aug 4 '13 at 10:05

1 Answer 1

up vote 14 down vote accepted

First we have $$a^2_{n}-a_{n+1}a_{n-1}=4$$ Proof: $$a_{n}(4a_{n-1})=a_{n-1}(4a_{n})$$ then we $$a_{n}(a_{n}+a_{n-2})=a_{n-1}(a_{n+1}+a_{n-1})$$ then $$a^2_{n}-a_{n+1}a_{n-1}=a^2_{n-1}-a_{n}a_{n-2}=\cdots=a^2_{2}-a_{3}a_{1}=4$$

so $$\cot^{-1}{a^2_{n}}=\cot^{-1}{\dfrac{a_{n}(4a_{n})}{4}}=\cot^{-1}{\dfrac{a_{n}(a_{n+1}+a_{n-1})}{a^2_{n}-a_{n+1}a_{n-1}}}=\cot^{-1}{\dfrac{a_{n+1}}{a_{n}}}-\cot^{-1}{\dfrac{a_{n}}{a_{n-1}}}$$ so $$\sum_{n=1}^{\infty}\cot^{-1}{a^2_{n}}=\cot^{-1}{(2+\sqrt{3})}=\dfrac{\pi}{12}$$ since $$\sum_{n=1}^{\infty}\cot^{-1}{a^2_{n}}=\cot^{-1}{a^2_{1}}+\lim_{N\to\infty}\sum_{n=2}^{N}\cot^{-1}{a^2_{n}}=\cot^{-1}{a^2_{1}}+\lim_{n\to\infty}\sum_{n=2}^{N}\left(\cot^{-1}{\dfrac{a_{n+1}}{a_{n}}}-\cot^{-1}{\dfrac{a_{n}}{a_{n-1}}}\right)$$ $$=\cot^{-1}{a^2_{1}}+\lim_{n\to\infty}\left(\cot^{-1}{\dfrac{a_{N+1}}{a_{N}}}-\cot^{-1}{\dfrac{a_{2}}{a_{1}}}\right)=\lim_{N\to\infty}\cot^{-1}{\dfrac{a_{N+1}}{a_{N}}}$$

let $a_{n+1}=4a_{n}-a_{n-1}$,we easy prove $$\lim_{n\to\infty}\dfrac{a_{n+1}}{a_{n}}=r>1$$is exst. so $$r^2=4r-1\Longrightarrow r=2+\sqrt{3}$$

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Whoa... I am speechless. Just beautiful. (+1) –  Prism Aug 4 '13 at 10:37
    
@math110, would you mind explaining the last step? My concern is how you are getting $2+\sqrt3$ in spite of the fact that as each $a_n$ is integer –  lab bhattacharjee Aug 4 '13 at 10:57
    
I have edit,@labbhattacharjee –  math110 Aug 4 '13 at 11:07
1  
What happened to the $\cot^{-1} a_1^2$ piece? –  Ron Gordon Aug 4 '13 at 11:10
    
$$\cot^{-1}{a^2_{1}}=\cot^{-1}{\dfrac{a_{2}}{a_{1}}}=\cot^{-1}{4}$$ –  math110 Aug 4 '13 at 11:12

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