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Consider a sequence $$ s_{n}=\sum_{r=1}^{n}\tan^{-1}\frac{2}{r^{2}} .$$

then we have

$$ \frac{1}{2}\sum_{r=1}^{n}\frac{2}{r^{2}} \leq s_{n} \ \leq \sum_{r=1}^{n}\frac{2}{r^{2}} \\\Rightarrow \frac{\pi^{2}}{6} \leq \lim_{n \to \infty}s_{n}\leq\frac{\pi^{2}}{3} $$

But how to attack to find the exact value of this limit? Anyone please?

Arctic Char
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Mathematics Man
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  • Also: https://math.stackexchange.com/q/738284/42969, https://math.stackexchange.com/q/3053616/42969 – Martin R Sep 03 '20 at 07:01
  • @MartinR Of the three links, the last one I consider to be most applicable and is the one I chose as the question this post duplicates. The first one, despite a superficial similarity (the only difference being $1/n^2$ instead of $2/n^2$), leads to a very different means of solution. – heropup Sep 03 '20 at 07:19

1 Answers1

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Here is a hint: $$\frac{2}{r^2} = \frac{2}{1 + (r^2 - 1)} = \frac{(r+1) - (r-1)}{1 + (r+1)(r-1)},$$ and $$\tan^{-1} \alpha - \tan^{-1} \beta = \tan^{-1} \frac{\alpha - \beta}{1 + \alpha \beta}.$$

heropup
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