Question

If $\sum _{n=1}^{\infty }2{\cot }^{-1}\left(\frac{n^2+n+4}{2}\right)=k\pi$ then find the value of k.

Answer 1

$\sum _{n=1}^{\infty }2{\cot }^{-1}(\frac{n^2+n+4}{2})=k\pi .......eq\left(1\right)$

${\cot }^{-1}(\frac{n^2+n+4}{2})$

We can rewrite above equation as.

$={\cot }^{-1}(\frac{1+\frac{n+1}{2}\times \frac{n}{2}}{\frac{n+1}{2}-\frac{n}{2}})$

$={\cot }^{-1}(\frac{n+1}{2})-{\cot }^{-1}(\frac{n}{2})......eq\left(2\right)$ $\because {\cot }^{-1}a-{\cot }^{-1}b={\cot }^{-1}(\frac{1+ab}{a-b})$

Here $a=\frac{n+1}{2}andb=\frac{n}{2}$

Put eq(2) in eq(1)

$2\sum _{n=1}^{\infty }{\cot }^{-1}(\frac{n+1}{2})-{\cot }^{-1}(\frac{n}{2})=k\pi .....eq\left(3\right)$

$forn=0$

$={\cot }^{-1}(\frac{1}{2})-{\cot }^{-1}(0)......eq\left(4\right)$

$forn=1$

$={\cot }^{-1}(\frac{2}{2})-{\cot }^{-1}(\frac{1}{2})......eq\left(5\right)$

$forn=3$

$={\cot }^{-1}(\frac{3}{2})-{\cot }^{-1}(\frac{2}{2})......eq\left(6\right)$

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$Forn=\infty$

$={\cot }^{-1}(\frac{\infty }{2})-{\cot }^{-1}(\frac{\infty -1}{2})......eq\left(7\right)$

Hence on adding eq(4) to eq(7) we get

$2\sum _{n=1}^{\infty }{\cot }^{-1}(\frac{n+1}{2})-{\cot }^{-1}(\frac{n}{2})=2(-{\cot }^{-1}\left(0\right)+{\cot }^{-1}\left(\infty \right))$

$2(-{\cot }^{-1}\left(0\right)+{\cot }^{-1}\left(\infty \right))=k\pi$

$=2(-\frac{\pi }{2}+0)=k\pi$

$⟹k\pi =-\pi$

Hence

$k=-1$