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[数列] $∑\cot^{-1}\frac{(n+1)^2}2$

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hbghlyj Posted 2023-8-13 22:53 |Read mode
cot(sum_(n=1)^∞ cot^(-1)(1/2 (1+n)^2))=0.3333333
$$\sum_{k=1}^{\infty} \operatorname{arccot}\left((1+k)^2\over2\right) = \operatorname{arccot}(\frac13)$$

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kuing Posted 2023-8-13 23:52
forum.php?mod=redirect&goto=findpost& … d=4922&pid=23275 是一样的,裂项解法见 4#

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