Sum Of An Infinite Special Series
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.
Select the correct option:
Solution
2
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.
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About This Question
- Subject
- mathematics
- Chapter
- sequence and series
- Topic
- sum of an infinite special series
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
2
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.
This hard difficulty mathematics question is from the chapter sequence and series, covering the topic of sum of an infinite special series. It appeared in the 2025 exam.
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