Question

Use the properties of infinite series to evaluate the following series. $$\sum_{k=0}^{\infty}\left[3\left(\frac{2}{5}\right)^{k}-2\left(\frac{5}{7}\right)^{k}\right]$$

Short answer

$$ Answer: The sum of the given infinite series is -2.

Step-by-step solution

Identify the two geometric series

The given series can be considered as a sum of two separate geometric series: $$\sum_{k=0}^{\infty}\left[3\left(\frac{2}{5}\right)^{k}-2\left(\frac{5}{7}\right)^{k}\right] = \sum_{k=0}^{\infty}3\left(\frac{2}{5}\right)^{k} - \sum_{k=0}^{\infty}2\left(\frac{5}{7}\right)^{k}.$$ The first geometric series is: $$\sum_{k=0}^{\infty}3\left(\frac{2}{5}\right)^{k}$$ with \(a_1 = 3\) and \(r_1 = \frac{2}{5}\). The second geometric series is: $$\sum_{k=0}^{\infty}2\left(\frac{5}{7}\right)^{k}$$ with \(a_2 = 2\) and \(r_2 = \frac{5}{7}\).

Determine the sum of each geometric series

For a convergent geometric series, the sum is given by the formula: \(S = \frac{a}{1-r}\), where \(a\) is the first term of the series and \(r\) is the common ratio. Since both \(r_1\) and \(r_2\) have an absolute value less than 1, both geometric series are convergent. For the first geometric series: $$S_1 = \frac{3}{1-\frac{2}{5}} = \frac{3}{\frac{3}{5}} = 5$$ For the second geometric series: $$S_2 = \frac{2}{1-\frac{5}{7}} = \frac{2}{\frac{2}{7}} = 7$$

Combine the sums of the geometric series

Now we can find the sum of the original series by subtracting the sum of the second geometric series from the sum of the first geometric series: $$\sum_{k=0}^{\infty}\left[3\left(\frac{2}{5}\right)^{k}-2\left(\frac{5}{7}\right)^{k}\right] = S_1 - S_2 = 5 - 7 = -2$$ Therefore, the overall sum of the given series is: $$\sum_{k=0}^{\infty}\left[3\left(\frac{2}{5}\right)^{k}-2\left(\frac{5}{7}\right)^{k}\right] = -2.$$

Key concepts

Infinite Series

When dealing with an infinite series, we address the sum of terms that continue indefinitely. A series is represented by a summation symbol and expresses the addition of an infinite number of terms. For instance, in the given exercise, the series runs from \(k=0\) to infinity. The terms within this series involve powers of a constant ratio \(r\), expressed as \(a_k(r)^k\), where \(a_k\) is a constant multiplier for each sequence.

Infinite series can either converge or diverge. Depending on the ratio of the terms as they progress to infinity, the series can reach a finite sum (converge) or not (diverge). It's fascinating how an infinite series like the sum in this exercise can have a relatively simple and finite result.

Series Convergence

For an infinite series to be useful, we often want it to "converge," meaning it approaches a specific value even as the number of terms becomes infinitely large. A geometric series is a common type of series where each term is a constant multiple \(r\) of the previous term.

To check for convergence in a geometric series, the absolute value of the ratio \(|r|\) must be less than 1. In the exercise, both series have ratios: \(r_1 = \frac{2}{5}\) and \(r_2 = \frac{5}{7}\), which are less than 1, confirming that they converge.

• This means, despite the series having an infinite number of terms, they sum up to a finite number.
• Convergence is crucial as it implies the series behaves in a predictable and measurable way.

Sum of a Series

The sum of a geometric series that converges can be calculated using a simple formula. If the first term of the series is \(a_1\) and the common ratio is \(r\), the sum \(S\) is given by \(S = \frac{a_1}{1-r}\) whenever \(|r| < 1\). This formula effectively summarizes infinitely many terms into a single finite result.

In the step by step solution, we see how the sum is calculated for both geometric series:

• First series: With \(a_1 = 3\) and \(r_1 = \frac{2}{5}\), its sum \(S_1\) is 5.
• Second series: With \(a_2 = 2\) and \(r_2 = \frac{5}{7}\), its sum \(S_2\) is 7.

These sums are then combined, letting us find the sum of the original series as \(-2\). Calculating the sum is not only about evaluating values but also about understanding how infinite elements can create a finite whole.