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If $$\sum_{n=1}^\infty cot^{-1}\left(2 + \frac{n(n+1)}{2}\right)=tan^{-1}a$$, then 'a' is equal to (A) 1 (B) 2 (C) 3 (D) 4

My aproach in such question is to break orignal function into difference of two functions using identies

$cot^{-1}(\frac{xy+1}{x-y}) = cot^{-1}x - cot^{-1}y$

In this $n^{2} + n + 4$ cannot be changed to xy+1

Is there any other way of doing it ?

Voneone
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  • Related : http://math.stackexchange.com/questions/193001/explicitly-finding-the-sum-of-arctan1-n2n1 – lab bhattacharjee Mar 28 '17 at 06:14
  • http://math.stackexchange.com/questions/415512/is-s-sum-r-1-infty-tan-1-frac2r2r2r4-finite and http://math.stackexchange.com/questions/144944/finding-tan-t-if-t-sum-tan-11-2t2/761847#761847 – lab bhattacharjee Apr 02 '17 at 10:48

1 Answers1

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HINT:

$\cot(A-B)=\dfrac{\cot A\cot B+1}{\cot B-\cot A}$

and $$\dfrac{4+n(n+1)}2=\dfrac{1+\dfrac n2\cdot\dfrac{n+1}2}{\dfrac{n+1}2-\dfrac n2}$$

lab bhattacharjee
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