 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/19/a/ii/solution

2/3

, so

\sum_{m = n}^{\infty} \frac{2 ^{m}}{3 ^{m}} = \frac{( 2/3 ) ^{n}}{1 - 2/3} = 3 (2/3)^{n} .

The outer geometric series therefore converges as well, with value

n = 0 \sum \infty m = n \sum \infty (2/3)^{m} = 3 n = 0 \sum \infty (2/3)^{n} = 9.

All summands are nonnegative, so there is no cancellation or conditional-convergence issue in evaluating the nested sums.

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