The sum of the trigonometric series...?

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August 13th 2008, 06:57 AM
fardeen_gen

The sum of the trigonometric series...?

The sum ∑(from n = 1 to ∞) arctan [n/(n^4 - 2n^2 + 2)] is equal to:
A) (arctan 2 + arctan 3)/4
B) 4. (arctan 1)
C) -π/16
D) 3π/16

More than one options may be correct.
August 14th 2008, 09:13 AM
fardeen_gen

To be more precise, it is http://alt1.artofproblemsolving.com/...7ebfe69b99.gif = ?
August 14th 2008, 10:09 AM
wingless

Quote:

Originally Posted by fardeen_gen

The sum ∑(from n = 1 to ∞) arctan [n/(n^4 - 2n^2 + 2)] is equal to:
A) (arctan 2 + arctan 3)/4
B) 4. (arctan 1)
C) -π/16
D) 3π/16

More than one options may be correct.

None of them is correct.

$\sum_{n=1}^\infty \arctan \frac{n}{n^4-2n^2+2} \approx 1.072215$

A) $\frac{\arctan 2 + \arctan 3}{4} \approx 0.58904$

B) $4 \arctan 1 = 4 \frac{\pi}{4} = \pi \approx 3.14159$

C) $-\frac{\pi}{16} \text{ is negative \dots}$

D) $\frac{3\pi}{16} \approx 0.58904$
August 15th 2008, 05:19 AM
fardeen_gen

One of the answers is definitely true - this was what my instructor told me. Are you sure wingless that the the given sum is not equal to any of the options??