 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/14/a/iii/solution

The absolutely convergent geometric series has sum

\sum_{k = 0}^{\infty} r^{k} = (1 - r)^{- 1}

for

0 < r < 1

. Its Cauchy product with itself is

(\sum_{k = 0}^{\infty} r^{k})^{2} = \sum_{m = 0}^{\infty} (\sum_{k = 0}^{m} 1) r^{m} = \sum_{m = 0}^{\infty} (m + 1) r^{m} .

There are exactly

m + 1

pairs of nonnegative integers whose sum is

m

. Relabeling

j = m + 1

proves

j = 1 \sum \infty j r^{j - 1} = \frac{1}{( 1 - r ) ^{2}} .

Absolute convergence justifies the multiplication and regrouping; no differentiation of an infinite series is needed.

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