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$$S = \arctan\frac{2}{1^2} + \arctan\frac{2}{2^2} + \arctan\frac{2}{3^2}+ \cdots+\arctan\frac{2}{n^2}$$

I tried to simplify this summation by substituting $\arctan(2/k^2)$ to $a_k$ so that$\ \tan(a_k) = 2/k^2$, then using the sum of tangent formula, but it didn't work. Can someone help me how to simplify this summation?

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Hint:

$$\dfrac2{n^2}=\dfrac{n+1-(n-1)}{1+(n-1)(n+1)}$$

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    $\begingroup$ You should add the not-well-known-formula-by-present-day-students $\tan(a-b)=\dfrac{\tan a - \tan b}{1+\tan a \tan b}$ $\endgroup$
    – Jean Marie
    Mar 11 '20 at 11:05
  • $\begingroup$ @JeanMarie, You are saying OP is trying to solve the current problem w/o knowing the basic formula :) $\endgroup$ Mar 11 '20 at 11:15
  • $\begingroup$ @lab Bhattacharjee Thank you, sir... What a unique IDEA!! Thank you for your generosity :) I think I will always transform $2/n^2 $ formula afterwards :)) $\endgroup$
    – Ivan_Raki
    Mar 11 '20 at 11:16
  • $\begingroup$ [+1] Besides for the formula... $\endgroup$
    – Jean Marie
    Mar 11 '20 at 11:17
  • $\begingroup$ @JeanMarie I'm so sorry for the lack of my knowledge, but I actually know the sum/difference formula of tangents, sir. I hope you don't look down on others again since there are a lot of intelligent people in the world :) $\endgroup$
    – Ivan_Raki
    Mar 11 '20 at 12:39

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