Sum Of An Infinite Special Series

Question

Determine the exact value of the convergent infinite sum whose kth term is k/2^k, taken over all positive integers k from one to \infty.

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This is an arithmetico-geometric series, and JEE Advanced sums such infinite series with the shift-and-subtract differencing technique. Let S = Σ_{k=1}^∞ k/2^k = 1/2 + 2/4 + 3/8 + .... Multiply by the ratio 1/2 to get S/2 = 1/4 + 2/8 + 3/16 + .... Subtracting, S - S/2 = 1/2 + 1/4 + 1/8 + ..., whose right side is a geometric series with first term 1/2 and ratio 1/2, summing to 1. Therefore S/2 = 1, giving S = 2. Option 1 mistakenly stops at the geometric sum without doubling. Option 3/2 truncates the series prematurely. Option 5/2 overcounts by misplacing a coefficient. Hence S = 2. This value also follows from the closed form Σ k x^k = x/(1 - x)^2 evaluated at x = 1/2, giving (1/2)/(1/4) = 2, which links the numerical series to the arithmetico-geometric generating function and offers an independent route JEE Advanced students should keep in reserve. Plausibility check: the partial sums 0.5, 1.0, 1.375, 1.625, 1.8125 climb steadily toward 2 without exceeding it, exactly the monotone-bounded behaviour expected of this convergent positive series.

Sum Of An Infinite Special Series

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