Problem 51 - Thomas Calculus

Question

Which series in Exercises \(49-68\) converge, and which diverge? Give reasons for your answers. If a series converges, find its sum. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{3}{2^{n}} $$

Step-by-Step Solution

Answer

The series converges by the Alternating Series Test and its sum is 1.

Step 1: Identify the Series

Given the series, it is: \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{3}{2^{n}} \). Recognize it as an alternating series because it has the form \((-1)^{n+1}\).

Step 2: Check for Alternating Series Test Conditions

For an alternating series \( \sum (-1)^{n} b_n \) to converge, where \( b_n = \frac{3}{2^{n}} \), the conditions are: 1. \( b_n > 0 \) for all \( n \).2. \( b_n \) should be decreasing.3. \( \lim_{n \to \infty} b_n = 0 \).Verify these conditions for \( b_n = \frac{3}{2^{n}} \).

Step 3: Verify Positivity

Check that \( \frac{3}{2^{n}} > 0 \) for all \( n \). This condition is satisfied because \( n \) starts at 1 and powers of 2 are positive.

Step 4: Verify Decreasing Nature

To show that \( \frac{3}{2^{n}} \) is decreasing, compare \( \frac{3}{2^{n}} \) and \( \frac{3}{2^{n+1}} \). Since \( \frac{3}{2^{n+1}} = \frac{3}{2} \times \frac{3}{2^{n}} \), \( \frac{3}{2^{n+1}} < \frac{3}{2^{n}} \). Hence, \( b_n \) is decreasing.

Step 5: Check the Limit Condition

Calculate \( \lim_{n \to \infty} \frac{3}{2^{n}} \). As \( n \) approaches infinity, \( \frac{3}{2^{n}} \to 0 \) since the exponential \( 2^n \) grows faster than the constant numerator 3.

Step 6: Apply Alternating Series Test Conclusion

Since all the conditions of the Alternating Series Test are satisfied, the series \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{3}{2^{n}} \) converges.

Step 7: Find the Sum of the Series

The series can be rewritten as an alternating geometric series with first term \( a = 3/2 \) and common ratio \( r = -1/2 \). The sum of an infinite geometric series \( S = \frac{a}{1 - r} \) if \( |r| < 1 \). Here, \( S = \frac{\frac{3}{2}}{1 - (-1/2)} = \frac{3}{2} \times \frac{2}{3} = 1 \).

Key Concepts

Series Convergence

Convergence in series can seem tricky, but it essentially addresses whether the sum of an infinite series can arrive at a finite value. **Series Convergence** specifically means that as we add more and more terms, the sum approaches a certain fixed number. When assessing convergence, one reliable tool is the Alternating Series Test. This test applies to series where consecutive terms keep changing sign, such as \( (-1)^{n+1} b_n \), which flips between adding and subtracting parts as you go down the terms. To determine if such a series converges, three conditions must be fulfilled:

• Each term's absolute value is positive: \( b_n > 0 \).
• The terms get smaller, or in other words, decrease: \( b_{n+1} < b_n \).
• The terms approach zero as \( n \) approaches infinity: \( \lim_{n \to \infty} b_n = 0 \).

If all these boxes are ticked, the series is deemed convergent!Checking these conditions shines light on whether we will have a converging sum. It’s essential as many real-world problems involve estimates related to infinite series, whether in physics, economics, or engineering.

Geometric Series

A geometric series is a type of series where each term is a constant multiple, called the common ratio, of the previous term. The series has a simple form: \( a + ar + ar^2 + ar^3 + \ldots \). For a geometric series to converge, the absolute value of the common ratio \( r \) must be less than 1. This means \( |r| < 1 \). When this condition is met, we can find the sum of the series using the formula:\[ S = \frac{a}{1 - r} \]where \( a \) is the first term, and \( r \) is the common ratio.In our exercise, once identified as alternating, the series \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{3}{2^{n}} \), turns into a geometric series with the first term \( a = \frac{3}{2} \) and common ratio \( r = -\frac{1}{2} \). This is valid since \( |-\frac{1}{2}| < 1 \), confirming convergence. It provides a vital shortcut to swiftly calculating the sum of complex series by recognizing the geometric form.

Limit of a Sequence

The concept of limits forms a fundamental part of understanding both sequences and series. A sequence is a list of numbers generated by some rule, and as you progress along this list (or as \( n \) goes to infinity), you might notice the numbers approaching a specific value. This specific value is called the **limit of the sequence**. For instance, in the series \( \sum_{n=1}^{\infty} \frac{3}{2^{n}} \), knowing that the limit of the sequence \( \frac{3}{2^{n}} \) as \( n \) tends to infinity is zero is critical. It’s because the powers of 2 in the denominator grow exponentially faster than the constant numerator. Hence, the fraction becomes smaller and smaller, eventually getting so close to zero that it practically reaches it, solidifying the series' capacity to converge.Understanding limits is essential not only in deciding the convergence of series but also in a wide array of mathematical analysis. They provide insight into the behavior of functions and sequences at extreme scales, giving us powerful tools in calculus and beyond.