Problem 43
Question
In Exercises \(43-58,\) use the Ratio Test to determine the convergence or divergence of the series. $$ \sum_{n=0}^{\infty} \frac{n !}{3^{n}} $$
Step-by-Step Solution
Verified Answer
The given series converges.
1Step 1: Recall the Ratio Test
The Ratio Test for series convergence states that, given a series \(\sum_{n=1}^{\infty} a_n\), if the limit \(L= \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\) exists, and if L < 1, then the series converges; if L > 1, then the series diverges, and if L = 1, the test is inconclusive.
2Step 2: Apply the Ratio Test to the Given Series
Let \(a_n= \frac{n !}{3^{n}}\). Then \(a_{n+1} = \frac{(n+1) !}{3^{n+1}}\). Calculate \(\lim_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right|\) which simplifies as \(\lim_{n \rightarrow \infty} \left( \frac{n !}{3^{n}} \cdot \frac{3^{n+1}}{(n+1) !}\right)\). The factorials and power of 3 should cancel out while simplification leading to \(\lim_{n\to\infty} \frac{3}{n+1}\).
3Step 3: Calculate the Limit
Evaluate the limit \(\lim_{n\to\infty} \frac{3}{n+1}\) which equals to 0.
4Step 4: Conclude with Ratio Test Result
Since 0 < 1, according to the Ratio Test, the given series \(\sum_{n=0}^{\infty} \frac{n !}{3^{n}}\) converges.
Key Concepts
Series ConvergenceInfinite SeriesFactorial NotationLimit Calculation
Series Convergence
Understanding series convergence is crucial when analyzing the behavior of infinite series, which are sequences of numbers added together ad infinitum. Think of it as a never-ending sum, where we wish to determine if there's a definitive value it approaches, or if it just keeps growing forever.
For the series \( \sum_{n=0}^{\infty} \frac{n !}{3^{n}} \) given in our exercise, we are interested in whether this particular series adds up to some finite number or not. If it does, we call the series convergent; if not, it's divergent. The Ratio Test is a specific method used to determine this convergence or divergence, and it's quite handy as it gives a clear-cut way to approach many series problems.
By applying the Ratio Test, if the limit of the ratio of consecutive terms is less than one, we can happily conclude that the series is convergent, meaning there's a finite sum out there that this infinite series is slowly inching towards.
For the series \( \sum_{n=0}^{\infty} \frac{n !}{3^{n}} \) given in our exercise, we are interested in whether this particular series adds up to some finite number or not. If it does, we call the series convergent; if not, it's divergent. The Ratio Test is a specific method used to determine this convergence or divergence, and it's quite handy as it gives a clear-cut way to approach many series problems.
By applying the Ratio Test, if the limit of the ratio of consecutive terms is less than one, we can happily conclude that the series is convergent, meaning there's a finite sum out there that this infinite series is slowly inching towards.
Infinite Series
An infinite series is a sum of an infinite sequence of terms. It's like a long train of numbers being added, extending indefinitely. While the concept may seem abstract, it offers powerful insight into functions and sequences in mathematics.
In our case, \( \sum_{n=0}^{\infty} \frac{n !}{3^{n}} \) represents an infinite series of fractions where the numerator is a factorial of the natural number \(n\) and the denominator is an exponential term with the base of three. The behavior of this infinite series can tell us whether we can associate a finite value to an otherwise endless procession of terms, which is quite a profound conclusion!
In our case, \( \sum_{n=0}^{\infty} \frac{n !}{3^{n}} \) represents an infinite series of fractions where the numerator is a factorial of the natural number \(n\) and the denominator is an exponential term with the base of three. The behavior of this infinite series can tell us whether we can associate a finite value to an otherwise endless procession of terms, which is quite a profound conclusion!
Factorial Notation
Factorial notation is a system used to multiply a series of descending natural numbers. Represented by the symbol \(n!\) where \(n\) is a non-negative integer, it means \(n \times (n-1) \times (n-2) \times \ldots \times 1\).
In our exercise, each term of the infinite series includes a factorial in the numerator. For instance, \(3!\) equals \(3 \times 2 \times 1\), or 6. Factorials grow extremely rapidly with increasing \(n\), which can significantly impact the terms of a series. However, in our Ratio Test calculation, the factorial in the numerator of a term largely cancels out with that of the next term, simplifying our problem considerably.
In our exercise, each term of the infinite series includes a factorial in the numerator. For instance, \(3!\) equals \(3 \times 2 \times 1\), or 6. Factorials grow extremely rapidly with increasing \(n\), which can significantly impact the terms of a series. However, in our Ratio Test calculation, the factorial in the numerator of a term largely cancels out with that of the next term, simplifying our problem considerably.
Limit Calculation
Limit calculation is a fundamental concept in calculus that determines the value a function approaches as the input approaches some value. In the context of series, we often look at the behavior of terms as \(n\) grows without bounds, or in other words, as \(n\) approaches infinity.
For the Ratio Test, our critical step is to calculate the limit of the ratio of consecutive terms as \(n\) goes to infinity, as noted in Step 3 of our solution steps. As seen in the exercise, the limit \(\lim_{n\to\infty} \frac{3}{n+1} = 0\), signaling that the ratio between successive terms shrinks to zero. As such, the value does not oscillate or diverge to infinity; it steadies out to a specific number, zero in this case, which tells us that the infinite series converges to a finite value.
For the Ratio Test, our critical step is to calculate the limit of the ratio of consecutive terms as \(n\) goes to infinity, as noted in Step 3 of our solution steps. As seen in the exercise, the limit \(\lim_{n\to\infty} \frac{3}{n+1} = 0\), signaling that the ratio between successive terms shrinks to zero. As such, the value does not oscillate or diverge to infinity; it steadies out to a specific number, zero in this case, which tells us that the infinite series converges to a finite value.
Other exercises in this chapter
Problem 43
In Exercises \(43-46,\) determine the degree of the Maclaurin polynomial required for the error in the approximation of the function at the indicated value of \
View solution Problem 43
In Exercises 43 and \(44,\) find a Maclaurin series for \(f(x)\). $$ f(x)=\int_{0}^{x}\left(e^{-t^{2}}-1\right) d t $$
View solution Problem 44
Find the sum of the convergent series. $$ \sum_{n=1}^{\infty}\left[(0.7)^{n}+(0.9)^{n}\right] $$
View solution Problem 44
Determine the convergence or divergence of the sequence with the given \(n\) th term. If the sequence converges, find its limit. \(a_{n}=\frac{(n-2) !}{n !}\)
View solution