递归数列 arctan之和

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如果数列$\an$满足$a_2=a_1+1=a_0+2$，且对于任意$n\ge2$，有如下递归关系$$a_{n-1}\left(a_{n}+a_{n+1}\right)+a_{n-2}\left(a_{n}-a_{n+1}\right)=\alpha+a_{n}^{2}+a_{n-1}^{2}$$求出
$$f(\alpha)=\sum_{n=1}^{\infty} \arctan\left(\frac{\alpha}{\alpha+2} \frac{1}{\left(a_{n}-a_{n-1}\right)^{2}}\right)$$
出自 数列极限技艺测题（地狱篇）.pdf (138.18 KB, Downloads: 52)

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hbghlyj

令$b_n=a_n-a_{n-1}$得$b_2=b_1=1,b_{n-1}b_{n+1}-b_n^2=\alpha$
\[f(\alpha)=\sum_{n=1}^{\infty} \arctan\left(\frac{\alpha}{\alpha+2}b_n^{-2}\right)\]
当$\alpha=0$时$b_n=1$, 显然$f(0)=0$
当$\alpha=1$时$b_n=\verb|Fibonacci[n]|$

1. N@Sum[ArcTan[1/(3 Fibonacci[n]^2)], {n, 1, 100}]/Pi

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输出$0.25$$$f(1)=\frac\pi4$$
当$\alpha=2$时$b_n=\frac{9+5 \sqrt{3}}{6}\left(2-\sqrt{3}\right)^n+\frac{9-5 \sqrt{3}}{6}\left(2+\sqrt3\right)^n$

1. N@Sum[ArcTan[1/(2(1/6 ((9-5 Sqrt[3]) (2+Sqrt[3])^n+(2-Sqrt[3])^n (9+5 Sqrt[3])))^2)],{n,1,200}]

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输出0.987246, 暂时没找到精确值
当$\alpha=-1$时$b_3=0,b_4=\infty$
当$\alpha=-2$时$b_n$为周期{1, 1, -1, -1}$\implies f(-2)=\sum_{n=1}^\infty\arctan\infty=\sum_{n=1}^\infty\frac\pi2=\infty$
当$\alpha=-3$时$b_n$为周期{1, 1, -2}, $\tan ^{-1}(3)+\tan ^{-1}(3)+\tan ^{-1}\left(\frac{3}{4}\right)=\pi\implies f(-3)=\infty$