Problem 32
Question
Determine whether the series is convergent, absolutely convergent, conditionally convergent, or divergent. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{2^{n}}{3 \cdot 5 \cdot 7 \cdots(2 n+1)}\)
Step-by-Step Solution
Verified Answer
The given series \(\sum_{n=1}^{\infty}(-1)^{n} \frac{2^{n}}{3 \cdot 5 \cdot 7 \cdots(2 n+1)}\) is absolutely convergent, based on the Ratio Test.
1Step 1: Apply the Ratio Test for Absolute Convergence
To apply the Ratio Test, we need to compute the limit of the ratio of consecutive absolute terms:
\[\lim_{n\to\infty}\frac{\left|\frac{(-1)^{n+1} 2^{n+1}}{3\cdot5\cdot7\cdots(2(n+1)+1)}\right|} {\left|\frac{(-1)^n 2^n}{3\cdot5\cdot7\cdots(2n+1)}\right|} \]
First, we simplify the expression inside the limit:
\[\frac{2^{n+1}}{3\cdot5\cdot7\cdots(2(n+1)+1)} \cdot \frac{3\cdot5\cdot7\cdots(2n+1)}{2^n} = \frac{2}{2n+3}\]
Now, evaluate the limit as n approaches infinity:
\[\lim_{n\to\infty}\frac{2}{2n+3} = 0\]
Since this limit is less than 1, the series is absolutely convergent.
2Step 2: Determine the Type of Convergence
Since the series is absolutely convergent, we know that it is convergent. Therefore, it is not necessary to test for conditional convergence or divergence.
3Step 3: Conclusion
The given series \(\sum_{n=1}^{\infty}(-1)^{n} \frac{2^{n}}{3 \cdot 5 \cdot 7 \cdots(2 n+1)}\) is absolutely convergent.
Key Concepts
Ratio TestAbsolutely ConvergentInfinite SeriesConditional Convergence
Ratio Test
The Ratio Test is a helpful tool for determining the convergence of a series. It's especially useful for series with terms involving powers or factorials. To apply the Ratio Test, we consider the absolute value of the ratio between consecutive terms of the series. We denote this by the formula:
\[L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]
In the exercise, the Ratio Test reveals that the limit \(L = 0\), which means the series converges absolutely. This test simplifies the process, giving a clear indication about the convergence without heavy computation.
\[L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]
- If \(L < 1\), the series converges absolutely.
- If \(L > 1\), the series is divergent.
- If \(L = 1\), the test is inconclusive.
In the exercise, the Ratio Test reveals that the limit \(L = 0\), which means the series converges absolutely. This test simplifies the process, giving a clear indication about the convergence without heavy computation.
Absolutely Convergent
A series is said to be absolutely convergent if the series of the absolute values of its terms converges. In simpler terms, if you take each term of the series, ignore its sign, and the resulting series sums to a finite number, then the original series is absolutely convergent.
Why is this important?
Why is this important?
- Absolute convergence implies convergence. If a series is absolutely convergent, it is also convergent in the standard sense.
- This property is extremely powerful, as it guarantees the behavior of the series regardless of the arrangement of terms.
- In the original exercise, using the Ratio Test, we determined that our series converges absolutely because the limit of the ratio test was less than 1.
Infinite Series
An infinite series is the sum of the terms of an infinite sequence. You can think of it as adding up endless numbers, one after another. Mathematically, it's represented as:
\[\sum_{n=1}^{\infty} a_n\]
Where each \(a_n\) represents a term in the series. The key question for any infinite series is whether it converges or diverges:
\[\sum_{n=1}^{\infty} a_n\]
Where each \(a_n\) represents a term in the series. The key question for any infinite series is whether it converges or diverges:
- Convergent: If adding all its terms will approach a specific value.
- Divergent: If the sum does not approach any unique value.
Conditional Convergence
Conditional convergence is a fascinating concept in the study of series. It occurs when a series is convergent, but not absolutely convergent.
In other words, if a series converges only when the terms are allowed to keep their sign, but not when converted to their absolute values, it is conditionally convergent.
In other words, if a series converges only when the terms are allowed to keep their sign, but not when converted to their absolute values, it is conditionally convergent.
- The most common example of a conditionally convergent series is the alternating harmonic series.
- This happens because there is a delicate balance between the signs and the magnitudes of terms that allows the series to approach a limit but would otherwise not stabilize.
Other exercises in this chapter
Problem 31
Determine whether the series is convergent, absolutely convergent, conditionally convergent, or divergent. \(\sum_{n=1}^{\infty}(-1)^{n-1} \frac{3 \cdot 5 \cdot
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Use the binomial series to find the power series representation of the function. Then find the radius of convergence of the series. \(f(x)=\frac{1}{\sqrt[3]{8+x
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Determine whether the series is convergent or divergent. \(\sum_{n=1}^{\infty} \frac{\tan ^{-1} n}{n^{3}+1}\)
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Determine whether the given series converges or diverges. If it converges, find its sum. \(\sum_{n=0}^{\infty} \frac{3^{n+1}}{5^{n}}\)
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