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Evaluate the double sum: sum from n=1 to infinity of (sum from k=1 to infinity of k / 2^(n+k)).

1. 1
2. 2
3. 3
4. 4

Correct answer: 2

Solution

The double sum sumₙ₌₁^(inf) sumₖ₌₁^(inf) k/2^(n+k) = sumₙ₌₁^(inf) (1/2ⁿ) * sumₖ₌₁^(inf) (k/2^k). First sum: sumₙ₌₁^(inf) (1/2)ⁿ = (1/2)/(1-1/2) = 1. Second sum: sumₖ₌₁^(inf) k*x^k = x/(1-x)² at x=1/2 => (1/2)/(1/2)² = (1/2)/(1/4) = 2. Product = 1 * 2 = 2.

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