∑n=1∞ tan−1n4−2n2+24n is equal to:

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^ { }_ { n=1 } tan^{-1}{4n}{n^4-2n^2+2} is equal to:

Question

$\sum_{n = 1}^{\infty} t a n^{- 1} \frac{4 n}{n ^{4} - 2 n ^{2} + 2}$ is equal to:

t a n^{- 1} 2 + t a n^{- 1} 3

4 t a n^{- 1} 1

π /2

se c^{- 1} (- 2)

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Mateo Discussed

\sum_{n = 1}^{\infty} t a n^{- 1} \frac{4 n}{n ^{4} - 2 n ^{2} + 2}

is equal to:

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Question Text

\sum_{n = 1}^{\infty} t a n^{- 1} \frac{4 n}{n ^{4} - 2 n ^{2} + 2}

is equal to: