Problem 37
Question
Determine the convergence or divergence of the sequence with the given \(n\) th term. If the sequence converges, find its limit. \(a_{n}=\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n-1)}{(2 n)^{n}}\)
Step-by-Step Solution
Verified Answer
The sequence diverges.
1Step 1: Identify the nth term and the (n+1)th term
The nth term provided in the exercise is \(a_{n}=\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n-1)}{(2 n)^{n}}\). To find the (n+1)th term, simply replace 'n' with 'n+1' to get \(a_{n+1}=\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n+1-1)}{(2 (n+1))^{n+1}} = \frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot 2n \cdot 2n+1}{(2n+2)^{n+1}}\).
2Step 2: Apply the Ratio Test
Using the Ratio Test to determine the convergence or divergence of the sequence involves dividing the (n+1)th term by the nth term, and taking the limit as n tends to infinity. \(\lim_{n\to\infty} \frac{a_{n+1}}{a_{n}} =\lim_{n\to\infty} \frac{\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot 2n \cdot 2n+1}{(2n+2)^{n+1}}}{\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n-1)}{(2 n)^{n}}} = \lim_{n\to\infty} \frac{(2n+1)(2n)^n}{(2n+2)^{n+1}}\).
3Step 3: Simplify the expression
Simplify the expression obtained in Step 2 to render its limit calculation easier. This gives \(\lim_{n\to\infty} \frac{(2n+1)}{(2n+2)}\).
4Step 4: Calculate the limit
Calculate the limit as n tends to infinity for \(\lim_{n\to\infty} \frac{(2n+1)}{(2n+2)}\). As both numerator and denominator tend to infinity, this is an indeterminate form of type \(\frac{\infty}{\infty}\). Dividing both numerator and denominator by n yields \(\lim_{n\to\infty} \frac{2+(1/n)}{2+(2/n)}\). Taking the limit as n tends to infinity gives the ratio \(\frac{2}{2} = 1\).
5Step 5: Determine convergence or divergence
As the value of the limit (ratio) is 1, which is not less than 1, so according to the Ratio Test, the sequence diverges.
Key Concepts
Ratio TestLimit CalculationIndeterminate FormsDivergence of Sequences
Ratio Test
The Ratio Test is a popular method for checking whether a series converges or diverges. It is especially helpful for series with terms that involve factorials, powers, or combinations thereof. Generally, the test involves:
- Finding the ratio of subsequent terms in the series, typically the (n+1)th term divided by the nth term.
- Taking the limit of this ratio as n approaches infinity.
- Checking the result of this limit:
- If the limit is less than 1, the series converges absolutely.
- If the limit is greater than 1, or infinite, the series diverges.
- If the limit equals 1, the test is inconclusive, and another method may be needed.
Limit Calculation
Limit calculation is a crucial skill in analyzing the convergence of sequences and series. It involves finding the value that a function or sequence approaches as the input approaches some value. In many cases, especially in sequence-related problems, the input approaches infinity.To perform a limit calculation:
- Simplify the expression as much as possible, often by canceling terms or factoring.
- Identify if the expression forms an indeterminate form like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
- If indeterminate, use tools like L'Hôpital's Rule or algebraic manipulation to resolve it.
Indeterminate Forms
Indeterminate forms are expressions that do not directly point to a specific value or conclusion. They occur often in calculus and often indicate the need for further manipulation to find a limit. Common forms include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \times \infty\), and more.
- In these situations, insights may be achieved by algebraic simplification or calculus-based theorems like L'Hôpital's Rule, which involves differentiating the numerator and denominator.
- Understanding these forms is vital in limit calculation processes.
Divergence of Sequences
The divergence of sequences indicates that the terms do not settle towards a single finite value as the sequence progresses to infinity. A sequence that diverges may increase infinitely, decrease without bound, or oscillate between values.Key indicators that a sequence diverges include:
- The terms increase or decrease without approaching a limit.
- Applying tests, like the Ratio Test, provide limits \(> 1\) or otherwise inconclusive results, such as equalling 1, which suggests non-convergence.
Other exercises in this chapter
Problem 36
Determine whether the series converges conditionally or absolutely, or diverges. $$ \sum_{n=0}^{\infty}(-1)^{n} e^{-n^{2}} $$
View solution Problem 37
Find the sum of the convergent series. $$ \sum_{n=0}^{\infty}\left(-\frac{1}{2}\right)^{n} $$
View solution Problem 37
In Exercises \(35-38,\) write an equivalent series with the index of summation beginning at \(n=1\). $$ \sum_{n=0}^{\infty} \frac{x^{2 n+1}}{(2 n+1) !} $$
View solution Problem 37
In Exercises \(35-38,\) use Taylor's Theorem to obtain an upper bound for the error of the approximation. Then calculate the exact value of the error. $$ \arcsi
View solution