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I am given $$S=\sum\limits_{n=1}^{23}\cot^{-1}\left(1+ \sum\limits_{k=1}^n 2k\right)$$

On expanding the sigma series becomes $$S= 23\cot^{-1}(3)+22\cot^{-1}(5) + \cdots + \cot^{-1}(47)$$ And in tan form as $$S= 23\tan^{-1}(1/3)+22\tan^{-1}(1/5) + \cdots + \tan^{-1}(1/47)$$ How to sum this series?

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  • $\begingroup$ Putting n = 1 , k can have one value . When I put n = 1 , k will have two values which will yield 2 values of arccot . This goes on until n = 23 $\endgroup$
    – Tesla
    Feb 24, 2015 at 10:11
  • $\begingroup$ Note that $\sum_{k=1}^n (2k) = 2 \frac{n(n+1)}{2} = n(n+1)$. So the sum must be: $S=cot^{-1}(3) + cot^{-1}(7) + ... + cot^{-1}(1+23*24)$ $\endgroup$
    – kryomaxim
    Feb 24, 2015 at 10:45
  • $\begingroup$ math.stackexchange.com/questions/193001/… $\endgroup$ Feb 25, 2015 at 17:54

1 Answer 1

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$$S=\sum\limits_{n=1}^{23}\cot^{-1}\left(1+ \sum\limits_{k=1}^n 2k\right)=\sum_{n=1}^{23}\cot^{-1}(1+n(n+1))=\sum_{n=1}^{23}\arctan\left(\frac{(n+1)-n}{1+n(n+1)}\right)=\sum_{n=1}^{23}[\arctan (n+1)-\arctan (n)]=\arctan(24)-\arctan(1)=\arctan\left(\frac{23}{25}\right)$$

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  • $\begingroup$ May I ask how did you do this step and the next one? $$\sum_{n=1}^{23}\arctan\left(\frac{(n+1)-n}{1+n(n+1)}\right)=\sum_{n=1}^{23}[ \arctan (n+1)-\arctan(n)]$$ $\endgroup$
    – rubik
    Feb 24, 2015 at 11:06
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    $\begingroup$ The fraction is $tan(\alpha -\beta)$ for $\alpha$ and $\beta$ with $tan \alpha =n+1$, $tan \beta =n$ @rubik. $\endgroup$
    – Fermat
    Feb 24, 2015 at 11:07
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    $\begingroup$ @Fermat: Ah wow, amazing. I never remember that identity. Thank you for the explanation! $\endgroup$
    – rubik
    Feb 24, 2015 at 12:29

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