Add up the infinite series 1/4 + 2/16 + 3/64 + …, whose k-th term is k × (1/4)^k. Express the total as a fraction in lowest terms.

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A weighted geometric sum

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Add up the infinite series $\frac{1}{4} + \frac{2}{16} + \frac{3}{64} + \dots$ , whose $k$ -th term is $k \times (\frac{1}{4})^{k}$ . Express the total as a fraction in lowest terms.

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Hint

Differentiate the geometric series $\sum x^{k}$ and multiply by $x$ .

Solution

For $∣ x ∣< 1$ , the sum of $x^{k}$ over $k \geq 0$ is $\frac{1}{1 - x}$ .
Differentiating and multiplying by $x$ gives $\sum k x^{k} = \frac{x}{( 1 - x ) ^{2}}$ .
At $x = \frac{1}{4}$ the squared factor is $\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$ .
So the sum is $\frac{\frac{1}{4}}{\frac{9}{16}} = \frac{4}{9}$ .

Answer: $\frac{4}{9}$

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