Problem 27
Question
\(19-40=\) Determine whether the series is absolutely convergent, conditionally convergent, or divergent. $$ \sum_{n=1}^{\infty} \frac{10^{n}}{(n+1) 4^{2 n+1}} $$
Step-by-Step Solution
Verified Answer
The series is absolutely convergent.
1Step 1: Rewrite the Series
The given series is \( \sum_{n=1}^{\infty} \frac{10^{n}}{(n+1) 4^{2n+1}} \). Let's simplify this expression:\[ \frac{10^{n}}{(n+1) 4^{2n+1}} = \frac{10^{n}}{(n+1) 4^{2n} \cdot 4} = \frac{10^{n}}{4}\cdot \frac{1}{4^{2n} (n+1)} = \frac{10^{n}}{4^{2n+1} (n+1)} \]
2Step 2: Test for Absolute Convergence
For absolute convergence, consider the series with absolute values:\[ \sum_{n=1}^{\infty} \left| \frac{10^{n}}{4^{2n+1} (n+1)} \right| \]Use the ratio test, which says a series \( \sum a_n \) converges absolutely if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \). Consider the ratio of consecutive terms:\[ \frac{a_{n+1}}{a_n} = \frac{\frac{10^{n+1}}{4^{2(n+1)+1}(n+2)}}{\frac{10^{n}}{4^{2n+1}(n+1)}} = \frac{10\cdot 10^{n}}{16 \cdot 4^{2n+1} (n+2)} \cdot \frac{4^{2n+1} (n+1)}{10^{n}} \]Simplify this:\[ = \frac{10 \cdot (n+1)}{16 \cdot (n+2)} = \frac{10(n+1)}{16(n+2)} \]
3Step 3: Evaluate the Limit for Ratio Test
Now, evaluate the limit as \( n \rightarrow \infty \):\[ \lim_{n \to \infty} \left| \frac{10(n+1)}{16(n+2)} \right| = \lim_{n \to \infty} \frac{10(n+1)}{16(n+2)} = \lim_{n \to \infty} \frac{10}{16} = \frac{5}{8} \]Since \( \frac{5}{8} < 1 \), the series converges absolutely.
Key Concepts
Absolute ConvergenceRatio TestInfinite Series
Absolute Convergence
The concept of absolute convergence is an essential part of understanding whether or not a series will converge. It determines if the sum of the absolute values of the series' terms converges.
For a series \( \sum a_n \), it is said to be absolutely convergent if the series of its absolute values \( \sum |a_n| \) converges.
When a series is absolutely convergent, it implies that the series converges regardless of the order of the terms. This property shows that absolute convergence is a stronger form of convergence than conditional convergence.
For a series \( \sum a_n \), it is said to be absolutely convergent if the series of its absolute values \( \sum |a_n| \) converges.
When a series is absolutely convergent, it implies that the series converges regardless of the order of the terms. This property shows that absolute convergence is a stronger form of convergence than conditional convergence.
- An absolutely convergent series always converges, but a converging series is not always absolutely convergent.
- If you can determine absolute convergence, you can affirm that the series converges in a very robust way.
- This concept is vital because it sometimes allows us to use various mathematical tools and techniques reliably.
Ratio Test
The ratio test is a popular method used to determine the absolute convergence of an infinite series. It's particularly useful for series with terms involving factorials or exponential expressions.
The ratio test involves looking at the ratio of consecutive terms in a series. For a series \( \sum a_n \), this test states:
This process is straightforward and particularly effective for complex series, making the ratio test a go-to tool for analysts and students alike.
The ratio test involves looking at the ratio of consecutive terms in a series. For a series \( \sum a_n \), this test states:
- Compute the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
- If \( L < 1 \), the series converges absolutely.
- If \( L > 1 \), the series diverges.
- If \( L = 1 \), the test is inconclusive, and other tests need to be used.
This process is straightforward and particularly effective for complex series, making the ratio test a go-to tool for analysts and students alike.
Infinite Series
An infinite series is a sum of an infinite sequence of terms. They are fundamental in calculus and analysis and can be represented as \( \sum_{n=1}^{\infty} a_n \).
Understanding infinite series involves grasping whether these series converge or diverge.
There are numerous tests and criteria, like the ratio test, to determine the nature of convergence of these series. Real-world applications of infinite series are seen in physics, engineering, and even finance, where approximations are necessary to make calculations feasible.
Infinite series allow us to understand and work with numbers and functions in ways that transcend simple arithmetic, bringing a higher level of insight into complex mathematical inquiries.
Understanding infinite series involves grasping whether these series converge or diverge.
- If a series converges, the sum approaches a finite number as more terms are added.
- Otherwise, if it diverges, the sum either increases indefinitely or oscillates without settling on a value.
There are numerous tests and criteria, like the ratio test, to determine the nature of convergence of these series. Real-world applications of infinite series are seen in physics, engineering, and even finance, where approximations are necessary to make calculations feasible.
Infinite series allow us to understand and work with numbers and functions in ways that transcend simple arithmetic, bringing a higher level of insight into complex mathematical inquiries.
Other exercises in this chapter
Problem 27
Is it possible to find a power series whose interval of convergence is \([0, \infty) ?\) Explain.
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Use a Maclaurin series in Table 1 to obtain the Maclaurin series for the given function. $$f(x)=\sin \pi x$$
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\(9-32\) n Determine whether the sequence converges or diverges. If it converges, find the limit. $$a_{n}=\left(1+\frac{2}{n}\right)^{n}$$
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Determine whether the series is convergent or divergent by expressing \(s_{n}\) as a telescoping sum (as in Example 6\() .\) If it is convergent, find its sum.
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