Question

State whether the given series converges and explain why. $$ 1-\sqrt{\frac{\pi}{3}}+\sqrt{\frac{\pi^{2}}{9}}-\sqrt{\frac{\pi^{3}}{27}}+\cdots $$

Short answer

The series converges by the Alternating Series Test.

Step-by-step solution

Identify the Series Type

The given series is an alternating series with terms involving square root and powers of \( \pi \). It can be expressed in the general form: \( a_n = (-1)^{n+1} \sqrt{\frac{\pi^n}{3^n}} \).

Rewrite the Terms

Each term of the series can be written as: \( a_n = \frac{(-1)^{n+1}\pi^{n/2}}{3^{n/2}} \). This rewrites each term to highlight the alternating nature and the geometric component.

Check the Convergence of the Alternating Series

According to the Alternating Series Test, an alternating series \( \sum (-1)^n a_n \) converges if \( a_{n+1} \le a_n \) for all \( n \) and \( \lim_{n \to \infty} a_n = 0 \).

Apply the Alternating Series Test

First, calculate \( a_n = \frac{\pi^{n/2}}{3^{n/2}} \). Notice that \( a_{n+1} = \frac{\pi^{(n+1)/2}}{3^{(n+1)/2}} \). Calculate \( \frac{a_{n+1}}{a_n} = \frac{\pi^{(n+1)/2}}{3^{(n+1)/2}} \times \frac{3^{n/2}}{\pi^{n/2}} = \left(\frac{\pi}{3}\right)^{1/2} \). Since \( \frac{\pi}{3} < 1 \), \( \left(\frac{\pi}{3}\right)^{1/2} < 1 \), ensuring that \( a_{n+1} \le a_n \). Also, \( \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{\pi^{n/2}}{3^{n/2}} = 0 \).

Conclusion on Convergence

Since both conditions for the Alternating Series Test are satisfied, the series \[ 1-\sqrt{\frac{\pi}{3}}+\sqrt{\frac{\pi^{2}}{9}}-\sqrt{\frac{\pi^{3}}{27}}+\cdots \] converges.

Key concepts

Alternating Series Test

The Alternating Series Test is a useful tool in determining the convergence of series that alternate between positive and negative terms. Consider a series of the form \( \sum (-1)^n a_n \) or \( \sum (-1)^{n+1} a_n \). In either case, to apply the Alternating Series Test, two conditions must be met:

• The absolute value of the terms \( a_n \) must decrease steadily as \( n \) increases. In mathematical terms, this means that for all \( n \), \( a_{n+1} \le a_n \).
• The limit of the terms \( a_n \) must approach zero as \( n \) approaches infinity, i.e., \( \lim_{n \to \infty} a_n = 0 \).

If both conditions hold true, the series converges. This test is particularly useful because it doesn't require us to find the sum of the series, only to determine its convergence.

Series Convergence

Series convergence is a fundamental concept in calculus. A series is said to converge if the sum of its infinite terms approaches a specific finite value.

• One basic criterion is the limit test, where if \( \lim_{n \to \infty} a_n eq 0 \), the series cannot converge.
• For an alternating series, we use specific tests like the Alternating Series Test to establish convergence.
• Sometimes, it's useful to first determine the type of series (geometric, harmonic, etc.) to decide on appropriate tests.

Understanding convergence is important because it determines whether a series can be used to approximate real values or functions.

Geometric Series

A geometric series is a series where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio.

• The general form of a geometric series is \( a + ar + ar^2 + ar^3 + \cdots \), where \( a \) is the first term and \( r \) is the common ratio.
• A geometric series converges if the absolute value of the common ratio is less than one, i.e. \( |r| < 1 \).
• The sum of an infinite geometric series can be calculated using the formula \( S = \frac{a}{1-r} \), provided \(|r| < 1 \).

Although the given exercise is an alternating series, it incorporates elements of a geometric series, as seen in the terms' construction \( \frac{(-1)^{n+1} \pi^{n/2}}{3^{n/2}} \), highlighting the significance of understanding geometric series for identifying and solving such problems.