Problem 7
Question
For Problems , determine whether the series converges or diverges. Explain your reasoning. $$ \frac{-1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\cdots-\frac{1}{n}-\cdots $$
Step-by-Step Solution
Verified Answer
The series \( \frac{-1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\cdots-\frac{1}{n}-\cdots \) converges according to the Alternating Series Test.
1Step 1: Identify the series
Identify the series \( \frac{-1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\cdots-\frac{1}{n}-\cdots \) as an alternating series since the signs of the terms alternate between positive and negative.
2Step 2: Determine the nth term
Determine the nth term of the series, which is \( -\frac{1}{n} \).
3Step 3: Apply the Alternating Series Test
Apply the Alternating Series Test for convergence. The test states that an alternating series converges if the sequences of absolute terms is monotonically decreasing and the limit as n goes to infinity of the absolute nth term is zero. -The sequence here is \( \frac{1}{n} \), and is monotonically decreasing since, as n increases, the terms get smaller. -The limit of the nth term as n goes to infinity is 0. \( \lim_{n \to \infty} \frac{1}{n} = 0 \). So the given series converges by the Alternating Series Test.
Key Concepts
Alternating Series TestConvergent SeriesDivergent Series
Alternating Series Test
The Alternating Series Test is a handy tool when dealing with series that change signs between terms. In simple words, an alternating series is a series whose terms alternate in sign. For instance, when you see a series like \( \frac{-1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\cdots \), you can spot that it alternates between negative and positive terms.
To use the Alternating Series Test for this kind of series, you must check two conditions:
The series in our problem meets these conditions and hence is convergent because of the Alternating Series Test.
To use the Alternating Series Test for this kind of series, you must check two conditions:
- The sequence of absolute terms \( b_n = \frac{1}{n} \) must be monotonically decreasing. This means as \( n \) grows larger, each term \( b_n \) should get smaller.
- The limit of \( b_n \) from this sequence as \( n \to \infty \) should be zero. In our case, \( \lim_{n \to \infty} \frac{1}{n} = 0 \).
The series in our problem meets these conditions and hence is convergent because of the Alternating Series Test.
Convergent Series
A convergent series is a series where the sum of its terms approaches a specific finite number. With each added term, the sum gets closer and closer to this number. It's like continually trying to hit a target, getting closer each time but never quite going beyond that target value.
To understand convergence better, consider our original series: \( \frac{-1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\cdots \). With the Alternating Series Test, we've confirmed it converges because each term is getting smaller and is approaching zero.
Convergent series are essential in calculus and mathematics because they ensure that infinite processes lead to finite results. This also tells us that even infinite sequences can sometimes "sum up" to a real number under certain conditions, such as alternating signs and decreasing term sizes as in this example.
To understand convergence better, consider our original series: \( \frac{-1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\cdots \). With the Alternating Series Test, we've confirmed it converges because each term is getting smaller and is approaching zero.
Convergent series are essential in calculus and mathematics because they ensure that infinite processes lead to finite results. This also tells us that even infinite sequences can sometimes "sum up" to a real number under certain conditions, such as alternating signs and decreasing term sizes as in this example.
Divergent Series
In contrast, a divergent series is one where the sum of its terms does not approach any specific number. As terms are added, the overall sum could increase indefinitely, decrease indefinitely, or simply alternate without settling to any limit. This lack of a finite sum means we cannot pinpoint a number it approaches.
For an illustrative divergence example, consider the harmonic series: \( 1 + \frac{1}{2} + \frac{1}{3} + \cdots \).Even though its terms also get smaller, they don't alternate, and the sum of this series diverges, which means it increases without bound.
Divergent series are quite different from convergent series. They remind us that not all infinite sums behave nicely. A series divergence implies that, despite possibly having diminishing terms, the sum still "escapes" reaching a finite result.
This understanding helps in clearly distinguishing between series that converge to a finite number from those that don't, playing a crucial role in mathematical analysis and applications.
For an illustrative divergence example, consider the harmonic series: \( 1 + \frac{1}{2} + \frac{1}{3} + \cdots \).Even though its terms also get smaller, they don't alternate, and the sum of this series diverges, which means it increases without bound.
Divergent series are quite different from convergent series. They remind us that not all infinite sums behave nicely. A series divergence implies that, despite possibly having diminishing terms, the sum still "escapes" reaching a finite result.
This understanding helps in clearly distinguishing between series that converge to a finite number from those that don't, playing a crucial role in mathematical analysis and applications.
Other exercises in this chapter
Problem 7
Make initial estimates to be sure that the answers you get are in the right ballpark. Suppose you borrow \(\$ 10,000\) at an interest rate of \(7 \%\) compounde
View solution Problem 7
Write the sum using summation notation. (a) \(\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+\cdots+\frac{2}{3^{n}}+\cdots\) (b) \(\frac{2}{3}+2+6+\cdots+2(3)^{n}+\cdots\
View solution Problem 8
Determine whether the series converges or diverges. If it converges, \(n d\) its sum. $$ \frac{3}{2}-\frac{3}{4}+\frac{3}{8}-\cdots+\frac{(-1)^{n+1} 3}{2^{n}}+\
View solution Problem 8
Make initial estimates to be sure that the answers you get are in the right ballpark. Suppose you borrow some money at an interest rate of \(6 \%\) compounded a
View solution