Problem 15
Question
Determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1} n !}{1 \cdot 3 \cdot 5 \cdot \cdot \cdot(2 n-1)} $$
Step-by-Step Solution
Verified Answer
The given series converges.
1Step 1: Apply the Ratio Test
The Ratio Test requires to find the limit as \( n \) approaches infinity of the absolute value of the ratio of the \( n+1 \) term to the \( n \) term of the series.\n Thus, apply the ratio test which states:\[\lim_{{n \to \infty}} \frac{a_{n+1}}{a_n} = L\]The terms of the series \( a_n \) and \( a_{n+1} \) are:\[a_n= \frac{(-1)^{n+1} n !}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2 n-1)}\]\[a_{n+1}= \frac{(-1)^{n+2} (n+1) !}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n+1)}\]
2Step 2: Calculate the limit
Divide \( a_{n+1} \) by \( a_n \) and apply limit as \( n \) approaches infinity.\[L= \lim_{n \to \infty} \left| \frac{(-1)^{n+2} (n+1) ! / 1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2 n+1)}{(-1)^{n+1} n ! / 1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2 n-1)}\right|\]Simplify\[L= \lim_{n \to \infty} \left| \frac{(-1)(n+1)(2n-1)!}{(2n+1)!}\right|\]\[L= \lim_{n \to \infty} \left| \frac{-1}{2n+1}\right|\]
3Step 3: Interpret the result
Since the absolute value of \( L \) is zero (which is less than 1), the given series converges by the Ratio Test. Hence, it results in an alternating series, each of whose terms is smaller than the one before.
Key Concepts
Ratio TestAlternating SeriesFactorial Expression
Ratio Test
The Ratio Test is a method in calculus used to determine whether a series converges or diverges. This test involves calculating the limit of the absolute value of the ratio of consecutive terms in the series as the number of terms goes to infinity.
The formal statement of the Ratio Test is:
The formal statement of the Ratio Test is:
- Given a series \(\sum_{n=1}^{\infty} a_n \), compute the limit \(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\).
- If \(L < 1\), the series converges.
- If \(L > 1\), or if \(L = \infty\), the series diverges.
- If \(L = 1\), the test is inconclusive.
Alternating Series
An alternating series is a series where the terms alternate in sign. This means positive and negative terms appear sequentially. The series in the original exercise is an example of an alternating series because of the factor \( (-1)^{n+1} \), which changes the sign of each successive term.
To determine the convergence of such a series, besides the Ratio Test, we can also use the Alternating Series Test. This test states that:
To determine the convergence of such a series, besides the Ratio Test, we can also use the Alternating Series Test. This test states that:
- For the series \(\sum (-1)^{n+1} b_n\), if the absolute value of terms \(b_n\) decrease steadily, i.e., \(b_{n+1} \leq b_n\), and \(\lim_{n \to \infty} b_n = 0\), then the series converges.
Factorial Expression
Factorial expressions may seem daunting, but they are simply products of an integer and all the integers before it.
For instance, the factorial notation \(n!\) means \(n \times (n-1) \times (n-2) \times ... \times 1 \). In the context of series, factorials often appear in sequences because they grow very fast, which affects the convergence of a series.
In the given example, the presence of factorials like \(n!\) and the product of odd numbers, such as \(1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1)\), creates a complex balance.
For instance, the factorial notation \(n!\) means \(n \times (n-1) \times (n-2) \times ... \times 1 \). In the context of series, factorials often appear in sequences because they grow very fast, which affects the convergence of a series.
In the given example, the presence of factorials like \(n!\) and the product of odd numbers, such as \(1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1)\), creates a complex balance.
- The denominator grows quickly, which makes each term smaller as \(n\) increases, assisting convergence.
- The numerator, which involves \(n!\), also grows but not as fast due to the odd factorial pattern on the denominator.
Other exercises in this chapter
Problem 15
In Exercises \(7-18\), find the Maclaurin polynomial of degree \(n\) for the function. $$ f(x)=\frac{1}{x+1}, \quad n=4 $$
View solution Problem 15
In Exercises 15-20, use the binomial series to find the Maclaurin series for the function. $$ f(x)=\frac{1}{(1+x)^{2}} $$
View solution Problem 15
Use the power series $$\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n} x^{n}$$ to determine a power series, centered at 0 , for the function. Identify the interval of
View solution Problem 16
Write the next two apparent terms of the sequence. Describe the pattern you used to find these terms. \(\frac{7}{2}, 4, \frac{9}{2}, 5, \ldots\)
View solution