"How to find value of this sum?∑m=1∞tan−1(2mm4+m2+2)" was answered by students like you

Jazmine Sweeney

Answered question

2022-04-20

How to find value of this sum?

\sum_{m = 1}^{\infty} \tan^{- 1} (\frac{2 m}{m^{4} + m^{2} + 2})

j3jell5

Beginner2022-04-21Added 17 answers

\frac{2 m}{m^{4} + m^{2} + 2} = \frac{2 m}{m^{4} + 2 m^{2} - m^{2} + 1 + 1}

= \frac{2 m}{{(m^{2} + 1)}^{2} - m^{2} + 1}

= \frac{2 m}{(m^{2} + m + 1) (m^{2} - m + 1) + 1}

= \frac{(m^{2} + m + 1) - (m^{2} - m + 1)}{(m^{2} + m + 1) (m^{2} - m + 1) + 1}

so almost done!

\arctan (a) - \arctan (b) = \arctan (\frac{a - b}{1 + a b})

\arctan (m^{2} + m + 1) - \arctan (m^{2} - m + 1) = \arctan (\frac{(m^{2} + m + 1) - (m^{2} - m + 1)}{(m^{2} + m + 1) (m^{2} - m + 1) + 1})

= \arctan \frac{2 m}{(m^{2} + m + 1) (m^{2} - m + 1) + 1}

now you have telescopic sumation, because

\arctan (m^{2} + m + 1) - \arctan (m^{2} - m + 1) = \arctan (m (m + 1) + 1) - \arctan (m (m - 1) + 1)

= f (m) - f (m - 1)

\sum_{m = 1}^{\infty} \tan^{- 1} (\frac{2 m}{m^{4} + m^{2} + 2})

= \sum_{m = 1}^{\infty} (f (m) - f (m - 1))

= lim_{m \to \infty} \tan^{- 1} (m^{2} + m + 1) - \tan^{- 1} 1

= \frac{π}{2} - \frac{π}{4} = \frac{π}{4}

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