Question

Find summation of cot^-1 (n^2 + n + 1) from 0 to infinity.

Answer 1

Solution!!

$\sf \displaystyle \sum\limits_{n=0}^{\infty }\;\cot ^{-1}(\sf n^{2}+n+1)$

$\sf T_{n}=\cot ^{-1}\;(n^{2}+n+1)$

$\sf =\tan ^{-1}\;\left(\dfrac{1}{n^{2}+n+1}\right)$

$\sf =\tan ^{-1}\;\left(\dfrac{(n+1)-n}{1+(n+1)n}\right)$

We know that $\sf \tan ^{-1}\;\left(\dfrac{x-y}{1+xy}\right)=\tan ^{-1}x-\tan ^{-1}y$

$\sf T_{n}=\tan ^{-1}(n+1)-\tan ^{-1}n$

$\sf T_{0}=\tan ^{-1}(1)-\tan ^{-1}(0)$

$\sf T_{1}=\tan ^{-1}(2)-\tan ^{-1}(1)$

$\sf T_{2}=\tan ^{-1}(3)-\tan ^{-1}(2)$

$\sf \vdots$

$\sf T_{n}=\tan ^{-1}(n+1)-\tan ^{-1}n$

$\sf T_{0}+T_{1}+T_{2}\dots T_{n}=\tan ^{-1}(n+1)-\tan ^{-1}(0)$

$\sf =\tan ^{-1}(n+1)$

$\sf T_{n}=\tan ^{-1}(n+1)$

$\sf n\xrightarrow{T_{n}}\ \infty =\tan ^{-1}(\infty +1)$

$\sf =\big {}^{ \pi }/{}_{ 2}$

$\boxed{\sf \displaystyle \sum\limits_{n=0}^{\infty }\;\cot ^{-1}(\sf n^{2}+n+1)={}^{\pi }/{}_{2}}$

Answer 2

Answer:

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