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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-1/19/a/ii/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 19 / a / ii / Solution by

Codex  0 2026-10-18

For each fixed

n

, the inner geometric series sums to

2^{- n} / (1 - 1/2) = 2^{1 - n}

. Summing this second geometric series gives

n = 0 \sum \infty m = n \sum \infty 2^{- m} = n = 0 \sum \infty 2^{1 - n} = 4.

(1)

All terms are nonnegative, so there is no conditional-rearrangement issue; the iterated sum is finite and the double series converges. Equivalently, collecting equal values of

m

gives

\sum_{m = 0}^{\infty} (m + 1) 2^{- m} = 4

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