Question

In Exercises 93-96, the terms of a series \(\sum_{n=1}^{\infty} a_{n}\) are defined recursively. Determine the convergence or divergence of the series. Explain your reasoning. $$ a_{1}=\frac{1}{2}, a_{n+1}=\frac{4 n-1}{3 n+2} a_{n} $$

Short answer

The series diverges as per the Ratio Test.

Step-by-step solution

Define the ratio of successive terms

Find the ratio of successive terms, which is \( a_{n+1}/a_{n}\) . For the given sequence this ratio is \( (4n-1)/(3n+2)\)

Apply the Ratio Test

We use the Ratio Test which states that a series \(\sum_{n=1}^{\infty} a_{n}\) converges if the limit \[L = \lim_{n \to \infty}\left| \frac{a_{n+1}}{a_{n}} \right|\] exists and is less than 1. Applying this formula, L becomes \[\lim_{n\rightarrow\infty}\frac{4n-1}{3n+2}\] assuming \( a_n \neq 0\).

Evaluate the limit

To find the limit above, we can divide each term by n, giving us \[\lim_{n\rightarrow\infty}\frac{4-1/n}{3+2/n}\]. As \( n \rightarrow \infty\) the terms with \( n \) in the denominator tend to zero, so the limit is \( 4/3\). Hence \( L = 4/3\).

Conclusion on the convergence or divergence

Since \( L = 4/3\) and this value is greater than 1, by the Ratio Test, the series diverges.

Key concepts

Ratio Test

The Ratio Test is a crucial tool for determining the convergence or divergence of infinite series. It's particularly useful when the terms of the series involve factorials, exponentials, or any form of recursive definitions. To perform this test, we compare the ratio of subsequent terms in the series.

Specifically, you consider the absolute value of the limit of the ratio of the (+1)th term to the nth term as n approaches infinity. The test states that if this limit, let's call it L, is less than 1, the series converges absolutely. If L is greater than 1, or if the limit does not exist, the series diverges. However, if L equals 1, the test is inconclusive, and we need to use other methods to determine convergence or divergence.

In the exercise provided, by finding the limit of the ratio \( L = \lim_{n \to \infty}\left| \frac{a_{n+1}}{a_{n}} \right| \) and determining that it is greater than 1 (specifically, \( 4/3 \) in this case), we can conclude that the series diverges, according to the Ratio Test.

Recursive Sequences

Recursive sequences are sequences defined using preceding terms. In a recursive sequence, the next term in the sequence is determined by applying a formula to the previous terms. This makes the series interdependent, as to calculate any particular term, one needs to know at least one, and often more, of the preceding terms.

In our textbook exercise, \( a_{1} \) is given as \( \frac{1}{2} \) and subsequent terms are given by the formula \( a_{n+1} = \frac{4 n-1}{3 n+2} a_{n} \). Such sequences can exhibit a variety of behaviors and understanding their progression is key to solving many problems in mathematics, including those related to series convergence.

Recursive sequences can sometimes be challenging to analyze, but certain tools, like the Ratio Test, can simplify the process by focusing on the relationship between consecutive terms rather than looking at the explicit formula for each term.

Limit of a Sequence

The limit of a sequence is a fundamental concept in calculus, representing the value that the terms of a sequence 'approach' as the index (usually n) becomes very large. If a sequence converges to a certain value, this means that as n increases, the terms get closer and closer to this limit.

To evaluate the limit of a recursively defined sequence like \( \lim_{n\rightarrow\infty}\frac{4n-1}{3n+2} \), we look at the behavior of the terms as n grows without bound. In mathematics, when faced with a complex fraction involving n, you may first simplify by dividing each term by n. This helps to isolate the highest powers of n, and as n goes to infinity, the lower order terms (those involving \( 1/n \) for example) tend to zero and can be neglected.

In the provided solution, this approach led to the conclusion that the limit of the ratio is \( 4/3 \), meaning that as n gets larger and larger, the terms of the sequence grow without bound, indicating that the sequence—and hence the series—diverges.