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Chapter 9: Problem 77

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

Expert verified
The series \(\sum_{n=1}^{\infty} a_n\) where \(a_{n+1}=\frac{4 n-1}{3 n+2} a_{n}\) and \(a_1 = 1/2\) converges because the ratio of consecutive terms converges to a value less than 1.

Step by step solution

01

Determine the first few terms of the series

To get a feel of this series, let's calculate the first few terms. We start with \(a_1 = 1/2\). Next to calculate \(a_2\), we substitute \(n = 1\) in the expression \(\frac{4n-1}{3n+2}a_n\), which gives \(\frac{3}{5}.\frac{1}{2} = 3/10\).Continuing in this way, we can find a few more terms. The individual terms are getting computed as \(a_3 = 7/24\), \(a_4 = 8/35\), \(a_5 = 33/154\) and so on.
02

Analyze the limiting behavior of the terms

It is observed that the terms \(a_n\) are getting smaller as \(n\) increases. This suggests that \(a_n\) tends towards zero as \(n\) goes to infinity.
03

Limit of Ratio of Consecutive Terms

To confirm the convergence of the series, we can use the ratio test. We will compute the limit of the ratios of consecutive terms:\[\lim_{n\to\infty} \frac{4n - 1}{3n + 2} = \lim_{n\to\infty} \frac{4 - 1/n}{3 + 2/n}\]In the limit as \(n\) goes to infinity, the terms with \(n\) in the denominator go to zero. Hence the limit of the ratio converges to \(4/3\). Since this ratio is less than 1, the series \(\sum_{n=1}^{\infty} a_n\) converges by the ratio test.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Recursive Sequences
A recursive sequence is a sequence in which each term after the first is defined as a function of the preceding terms. To better understand recursive sequences, it's helpful to look at a simple example, such as the one given in our exercise with the initial term given by \( a_1 = \frac{1}{2} \) and the recursive formula \( a_{n+1} = \frac{4n - 1}{3n + 2} \cdot a_n \).

Starting with \( a_1 \), we use the formula to find each subsequent term. This sequence forms the basis for determining the behavior of the series to which these terms contribute. As we noticed, the terms tend to get smaller, suggesting they may approach zero. This behavior is a crucial first indicator when analyzing the series鈥� convergence.
Limit of a Sequence
The limit of a sequence is the value that the terms of a sequence approach as the index (typically \( n \)) goes to infinity. If a sequence has a limit, then it can be said to converge to that value. For example, in our exercise, we observe that the terms of the sequence \( a_n \) appear to be getting smaller as \( n \) increases, indicating that they may have a limit of zero.

To determine the limit of a sequence rigorously, we often look for patterns or apply known limits, such as \( \lim_{n \to \infty} \frac{1}{n} = 0 \). In some cases, we may apply operations like factorization or rationalization to make the limit more apparent. The concept of limits is central to many areas of calculus and analysis, especially when determining the convergence of series.
Ratio Test
The ratio test is an essential tool in determining the convergence of series, especially when terms are defined recursively. To apply the ratio test, we compute the limit of the ratio of consecutive terms \( \lim_{n \to \infty} \frac{a_{n+1}}{a_n} \). If this limit is less than 1, the series converges; if it is greater than 1, the series diverges; and if it equals 1, the test is inconclusive.

In our exercise, the limit of the ratio of consecutive terms is \( \lim_{n \to \infty} \frac{4n - 1}{3n + 2} \), which simplifies to \( \frac{4}{3} \). Since this is greater than 1, it initially suggests divergence, but there was likely a misstep here; the ratio should be less than 1 for the series to converge (as typically expected in a convergent geometric series). To avoid confusion, ensure that the limit calculation correctly evaluates to a ratio less than 1 when applying the ratio test for convergence. This kind of careful analysis is a cornerstone for correctly assessing the behavior of series in mathematical studies.

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Most popular questions from this chapter

(a) Let \(f(x)=\sin x\) and \(a_{n}=n \sin 1 / n\). Show that \(\lim _{n \rightarrow \infty} a_{n}-f^{\prime}(0)=1\) (b) Let \(f(x)\) be differentiable on the interval \([0,1]\) and \(f(0)=0 .\) Consider the sequence \(\left\\{a_{n}\right\\}\), where \(a_{n}=n f(1 / n) .\) Show that \(\lim _{n \rightarrow \infty} a_{n}=f^{\prime}(0)\).

Investigation Consider the function \(f\) defined by \(f(x)=\left\\{\begin{array}{ll}e^{-1 / x^{2}}, & x \neq 0 \\ 0, & x=0\end{array}\right.\) (a) Sketch a graph of the function. (b) Use the alternative form of the definition of the derivative (Section 2.1) and L'H么pital's Rule to show that \(f^{\prime}(0)=0\). [By continuing this process, it can be shown that \(f^{(n)}(0)=\) 0 for \(n>1 .]\) (c) Using the result in part (b), find the Maclaurin series for \(f\). Does the series converge to \(f ?\)

determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. $$ \begin{aligned} &\text { If } f(x)=\sum_{n=0}^{\infty} a_{n} x^{n} \text { converges for }|x|<2, \text { then }\\\ &\int_{0}^{1} f(x) d x=\sum_{n=0}^{\infty} \frac{a_{n}}{n+1} \end{aligned} $$

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