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Evaluate: lim (n->infinity) tan[ sum from r=1 to n of arctan(1 / (1 + r + r²)) ].

1. 1
2. 0
3. -1
4. 2

Correct answer: 1

Solution

Using the identity arctan(a) - arctan(b) = arctan((a-b)/(1+ab)), we get arctan(1/(1+r+r²)) = arctan(r+1) - arctan(r). The sum telescopes: Sₙ = arctan(n+1) - arctan(1). As n->infinity, arctan(n+1)->pi/2. So S_infinity = pi/2 - pi/4 = pi/4. Therefore tan(pi/4) = 1.

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