The value of S = ∑ _ n = 1 ^ ∞ ^ -1 2n n^ 4 + n^ 2 + 2 is equal to

Given, $S = \mathop{∑}_{n = 1}^{\infty}\tan^{- 1}\dfrac{2n}{{n}^{4} + {n}^{2} + 2} . . . \left(i\right)$

Let ${n}^{4} + {n}^{2} + 1 = \left[{\left({n}^{2}\right)}^{2} + {1}^{2} + 2\left({n}^{2}\right)\left(1\right)\right] - {n}^{2}$

$= {\left({n}^{2} + 1\right)}^{2} - {n}^{2}$

$= \left({n}^{2} + n + 1\right)\left({n}^{2} - n + 1\right) . . . \left(ii\right)$

Let ${u}_{n} = \tan^{- 1}\left(\dfrac{2n}{{n}^{4} + {n}^{2} + 2}\right)$

$= \tan^{- 1}\dfrac{2n}{1 + \left({n}^{4} + {n}^{2} + 1\right)}$

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

${u}_{n} = \tan^{- 1}\left({n}^{2} + n + 1\right) - \tan^{- 1}\left({n}^{2} - n + 1\right) . . . \left(iii\right)$

On putting $n = 1, 2, 3, \ldots \ldots$ successively in Eq. (iii), we get

${u}_{1} = \tan^{- 1}3 - \tan^{- 1}1$

${u}_{2} = \tan^{- 1}7 - \tan^{- 1}3$

${u}_{3} = \tan^{- 1}13 - \tan^{- 1}7$

${u}_{n} = \tan^{- 1}\left({n}^{2} + n + 1\right) - \tan^{- 1}\left({n}^{2} - n + 1\right)$

On adding vertically, we get

$\mathop{∑}_{n = 1}^{\infty}{u}_{n} = \tan^{- 1}\left({n}^{2} + n + 1\right) - \tan^{- 1}1$

$S = \lim_{n \to \infty}\mathop{∑}_{n = 1}^{\infty}{u}_{n}$ [from Eq. (i)]

$= \lim_{n \to \infty}\tan^{- 1}\left({n}^{2} + n + 1\right) - \tan^{- 1}1$

$= \dfrac{\pi}{2} - \dfrac{\pi}{4} = \dfrac{\pi}{4}$