Problem 46
Question
Let \( \sum a_n \) be a series with positive terms and let \( r_n = a_{n+1} / a_n. \) Suppose that \( lim_{n \to \infty} r_n = L < 1, \) so \( \sum a_n \) converges by the Ratio Test. As usual, we let \( R_n \) be the remainder after \( n \) terms, that is, \( R_n = a_{n+1} + a_{n+2} + a_{n+3} + \cdot \cdot \cdot \) (a) If \( \\{r_n\\} \) is a decreasing sequence and \( r_{n+1} < 1, \) show, by summing a geometric series, that \( R_n \le \frac {a_{n+1}}{1 - r_{n+1}} \) (b) If \( \\{r_n\\} \) is a decreasing sequence, show that \( R_n \le \frac {a_{n+1}}{1 - L} \)
Step-by-Step Solution
Verified Answer
(a) and (b) are proven by summing a geometric series using the Ratio Test conditions.
1Step 1: Understanding the given
You are given a series with positive terms \( \sum a_n \), and the ratio \( r_n = \frac{a_{n+1}}{a_n} \) approaches a limit \( L < 1 \) as \( n \to \infty \). This means \( \sum a_n \) converges by the Ratio Test. Also, \( R_n \) is defined as the remainder of the series after \( n \) terms.
2Step 2: Part (a) restatement
You need to show that if \( \{ r_n \} \) is a decreasing sequence and \( r_{n+1} < 1 \), then \( R_n \leq \frac{a_{n+1}}{1 - r_{n+1}} \) by considering it as a geometric series.
3Step 3: Recognizing a geometric series
Given \( r_n \) is decreasing and \( r_{n+1} < 1 \), every term in the series of remainders \( R_n = a_{n+1} + a_{n+2} + ... \) can be expressed using the ratio: \( a_{n+k} = a_{n+1} r_{n+1}^{k-1} \). Thus, \( R_n = a_{n+1}(1 + r_{n+1} + r_{n+1}^2 + ...) \), which is a geometric series with first term \( a_{n+1} \) and ratio \( r_{n+1} \).
4Step 4: Sum of geometric series
The sum of an infinite geometric series with first term \( a_{n+1} \) and ratio \( r_{n+1} \) where \( |r_{n+1}| < 1 \) is \( \frac{a_{n+1}}{1 - r_{n+1}} \). Therefore, \( R_n = a_{n+1}(1 + r_{n+1} + r_{n+1}^2 + ...) = \frac{a_{n+1}}{1 - r_{n+1}} \), proving the required inequality: \( R_n \leq \frac{a_{n+1}}{1 - r_{n+1}} \).
5Step 5: Part (b) restatement
Now, show that \( R_n \leq \frac{a_{n+1}}{1 - L} \) given that \( \{ r_n \} \) is a decreasing sequence.
6Step 6: Applying the limit condition
Since \( \{ r_n \} \) is decreasing and \( \lim_{n \to \infty} r_n = L < 1 \), for large \( n \), \( r_{n+1} \) is approximately \( L \) and \( r_{n+1} \leq L \). Therefore, we use \( r_{n+1} < 1 \) and the relation derived in Step 4: \( R_n \leq \frac{a_{n+1}}{1 - r_{n+1}} \) and since \( r_{n+1} \leq L \), it follows that \( R_n \leq \frac{a_{n+1}}{1 - L} \) as well.
Key Concepts
Ratio TestGeometric SeriesLimit of a SequenceRemainder of a Series
Ratio Test
The Ratio Test is a popular method for testing the convergence of a series with positive terms. This test uses the idea of comparing each term of a series to its next term. Specifically, for a series \( \sum a_n \), the ratio \( r_n = \frac{a_{n+1}}{a_n} \) is considered. If the limit of \( r_n \) as \( n \to \infty \) is less than 1, denoted as \( L < 1 \), then the series converges.
This is because the terms of the series decrease rapidly enough to sum to a finite total. Think of it as each term getting smaller and smaller, reducing the chances of the sum blowing up to infinity.
This is because the terms of the series decrease rapidly enough to sum to a finite total. Think of it as each term getting smaller and smaller, reducing the chances of the sum blowing up to infinity.
- If \( L < 1 \), the series converges absolutely.
- If \( L > 1 \), or the limit does not exist, the series may diverge.
- If \( L = 1 \), the test is inconclusive.
Geometric Series
A geometric series is one of the simplest types of series and has terms that form a geometric progression. In a geometric series, each term is a multiple of the previous term by a constant factor, known as the ratio \( r \).
The formula for the sum of an infinite geometric series is \( \frac{a}{1-r} \), where \( a \) is the first term and \( |r| < 1 \). This formula is derived from repeatedly multiplying by \( r \) and summing the series until its value stabilizes as \( n \) becomes very large.
For example:
The formula for the sum of an infinite geometric series is \( \frac{a}{1-r} \), where \( a \) is the first term and \( |r| < 1 \). This formula is derived from repeatedly multiplying by \( r \) and summing the series until its value stabilizes as \( n \) becomes very large.
For example:
- If you have a series \( 2 + 2\cdot0.5 + 2\cdot0.5^2 + \ldots \), it is geometric with \( a = 2 \) and \( r = 0.5 \).
- Such a series converges to \( \frac{2}{1-0.5} = 4 \).
Limit of a Sequence
The limit of a sequence is a fundamental concept in calculus. For a sequence of numbers \( \{ r_n \} \), the limit \( \lim_{n \to \infty} r_n = L \) tells us what value the sequence approaches as \( n \) becomes infinitely large.
In the context of series convergence, understanding limits is crucial. When the limit of the ratio of consecutive terms \( r_n \) exists and is less than 1, we apply the Ratio Test to conclude the series' convergence. Knowing the limit allows us to predict the behavior of a series' future terms reliably.
In the context of series convergence, understanding limits is crucial. When the limit of the ratio of consecutive terms \( r_n \) exists and is less than 1, we apply the Ratio Test to conclude the series' convergence. Knowing the limit allows us to predict the behavior of a series' future terms reliably.
- If \( L < 1 \), the terms decrease quickly, leading to convergence.
- If \( L > 1 \), the sequence grows, causing divergence.
- The closer \( L \) is to zero, the faster the terms of the series shrink.
Remainder of a Series
The remainder of a series \( R_n \) represents the sum of the series' terms that remain after the first \( n \) terms have been considered. It is defined as:\[R_n = a_{n+1} + a_{n+2} + a_{n+3} + \cdots\]
The remainder is vital in approximating the sum of the series. As \( n \) increases, \( R_n \) usually becomes very small if the series converges. Using tools such as the geometric series sum gives us estimates of \( R_n \), ensuring our approximations of the total sum are accurate.
The remainder is vital in approximating the sum of the series. As \( n \) increases, \( R_n \) usually becomes very small if the series converges. Using tools such as the geometric series sum gives us estimates of \( R_n \), ensuring our approximations of the total sum are accurate.
- In the exercise, it is shown that \( R_n \leq \frac{a_{n+1}}{1 - r_{n+1}} \) if each ratio \( \{ r_n \} \) is decreasing and less than 1.
- This bounding helps to assure that the largest possible error in our sum is restricted.
Other exercises in this chapter
Problem 45
Determine whether the sequence converges or diverges. If it converges, find the limit. \( a_n = n \sin (1/n) \)
View solution Problem 46
Find the Maclaurin series of \( f \) (by any method) and its radius of convergence. Graph \( f \) and its first few Taylor polynomials on the same screen. What
View solution Problem 46
If \( \sum a_n \) and \( \sum b_n \) are both convergent series with positive terms, is it true that \( \sum a_n b_n \) is also convergent?
View solution Problem 46
Find all values of \( c \) for which the following series converges. \( \displaystyle \sum_{n = 1}^{\infty} \left( \frac {c}{n} - \frac {1}{n + 1} \right) \)
View solution