Question

Use the properties of infinite series to evaluate the following series. $$\sum_{k=1}^{\infty} \frac{4}{12^{k}}$$

Short answer

Answer: The sum of the given infinite geometric series is $\frac{4}{11}$.

Step-by-step solution

Determine the common ratio

The given series is $$\sum_{k=1}^{\infty} \frac{4}{12^k}$$ Let's identify the common ratio \(r\). In a geometric series, the ratio between consecutive terms is constant. Thus, we can find the common ratio by dividing the second term by the first term: $$r = \frac{\frac{4}{12^2}}{\frac{4}{12^1}} = \frac{1}{12}$$ The common ratio is \(\frac{1}{12}\).

Check for convergence

For an infinite geometric series to converge, the absolute value of the common ratio must be less than 1: $$|r| < 1$$ In this case, \(|\frac{1}{12}| < 1\), so the series converges.

Apply the sum formula

Since we know the series converges, we can apply the infinite geometric series sum formula: $$S = \frac{a}{1-r}$$ where \(S\) is the sum of the series, \(a\) is the first term, and \(r\) is the common ratio. For our series, the first term \(a = \frac{4}{12^1} = \frac{1}{3}\), and the common ratio \(r = \frac{1}{12}\). Plugging these values into the formula, we get: $$S = \frac{\frac{1}{3}}{1-\frac{1}{12}}$$ To simplify the expression, let's find a common denominator for the fractions inside the parenthesis. Since 3 and 12 both divide by 12, let's convert both fractions to have a denominator of 12: $$S = \frac{\frac{4}{12}}{1-\frac{1}{12}} = \frac{\frac{4}{12}}{\frac{11}{12}}$$

Simplify the expression

To simplify the expression for the sum, we can divide the fractions by multiplying the numerator by the reciprocal of the denominator: $$S = \frac{4}{12} \cdot \frac{12}{11}$$ The factor of 12 in the numerator and denominator can be canceled: $$S = \frac{4}{11}$$ So, the sum of the given infinite geometric series is \(\frac{4}{11}\).