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Sum Of An Infinite Special Series

Hardmathematics

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:

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About This Question

Subject
mathematics
Chapter
sequence and series
Topic
sum of an infinite special series
Difficulty
Hard
Year
2025
Tags
advanced-calculus-drillarithmetico-geometric-seriesshift-and-subtractinfinite-sumconvergence

Solution

Correct Answer:

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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