Problem 20
Question
Verify that the infinite series converges. $$ \sum_{n=1}^{\infty} \frac{1}{n(n+2)}(\text { Use partial fractions. }) $$
Step-by-Step Solution
Verified Answer
The series \( \sum_{n=1}^{\infty} \frac{1}{n(n+2)} \) converges to 1.5.
1Step 1: Express the term as a sum of partial fractions
First, we need to express \( \frac{1}{n(n+2)} \) as a sum of partial fractions. To do this, we consider that \( \frac{1}{n(n+2)} \) should be equal to \( \frac{A}{n} + \frac{B}{n+2} \). By solving this equation we find that A = 1 and B = -1.
2Step 2: Rewrite the series with simpler terms
Next, by substituting A and B into the series, we get \( \sum_{n=1}^{\infty} ( \frac{1}{n} - \frac{1}{n+2} ) \). This is easier to handle since we have broken the terms into simpler fractions.
3Step 3: Performing the summation
Then, we perform the summation for the first few terms, which is given by \( (1 - \frac{1}{3}) + (\frac{1}{2} - \frac{1}{4}) + (\frac{1}{3} - \frac{1}{5}) + ... \). We can notice that many terms cancel each other out: -1/3 cancels with the 1/3 in the next term, -1/4 cancels with the 1/4 in the following term, and so on. This continues for all terms.
4Step 4: Determining the convergence
Finally, as all the terms cancel each other out, the series converges to \( 1 + \frac{1}{2} \), which equals to 1.5.
Key Concepts
Partial FractionsSeries SummationConvergence Tests
Partial Fractions
Understanding partial fractions is essential when dealing with complex rational expressions, especially in calculus when we work with integration or series. When faced with a fraction that has a polynomial in the denominator, we often decompose it into a sum of simpler fractions, each of which has a denominator that is a factor of the original polynomial.
For example, the term \( \frac{1}{n(n+2)} \) can be split into two simpler fractions, \( \frac{A}{n} \) and \( \frac{B}{n+2} \) because \( n(n+2) \) can be factored into \( n * (n+2) \). To find the values of \( A \) and \( B \) that make this equation true, we solve for these constants. Once we have determined that \( A = 1 \) and \( B = -1 \) in our example, we can rewrite the original complex fraction as \( \frac{1}{n} - \frac{1}{n+2} \), greatly simplifying the problem at hand.
For example, the term \( \frac{1}{n(n+2)} \) can be split into two simpler fractions, \( \frac{A}{n} \) and \( \frac{B}{n+2} \) because \( n(n+2) \) can be factored into \( n * (n+2) \). To find the values of \( A \) and \( B \) that make this equation true, we solve for these constants. Once we have determined that \( A = 1 \) and \( B = -1 \) in our example, we can rewrite the original complex fraction as \( \frac{1}{n} - \frac{1}{n+2} \), greatly simplifying the problem at hand.
Series Summation
Series summation involves adding up the terms of a sequence to find the total or a partial sum. In the context of infinite series, we're interested in what happens as we add more and more terms; does the total converge to a finite value, or does it grow without bound?
When we use partial fractions to rewrite complex terms as simpler ones, we often encounter series that are easier to work with. After performing partial fraction decomposition on \( \frac{1}{n(n+2)} \) and rewriting it as \( (\frac{1}{n} - \frac{1}{n+2}) \), we notice a pattern called 'telescoping.' In a telescoping series, many terms cancel out when we write out the sequence of partial sums, just as in our rewritten series: \( 1 - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \ldots \). This ultimately simplifies the task of finding the sum of an infinite series.
When we use partial fractions to rewrite complex terms as simpler ones, we often encounter series that are easier to work with. After performing partial fraction decomposition on \( \frac{1}{n(n+2)} \) and rewriting it as \( (\frac{1}{n} - \frac{1}{n+2}) \), we notice a pattern called 'telescoping.' In a telescoping series, many terms cancel out when we write out the sequence of partial sums, just as in our rewritten series: \( 1 - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \ldots \). This ultimately simplifies the task of finding the sum of an infinite series.
Convergence Tests
Infinite series convergence is a key concept in calculus, where we determine whether an infinite series has a finite limit or not. Convergence tests are tools to conclude about the behavior of an infinite series. Common tests include the Ratio Test, Root Test, and the Integral Test, among others. For the series we're considering here, \( \sum_{n=1}^{\infty} \frac{1}{n(n+2)} \), the behavior of the series is easy to predict due to its telescoping nature.
Applying the knowledge that successive terms cancel each other out, we recognize that the series converges to a finite number, because only a few terms of the series do not cancel and contribute to the final sum. Specifically, after cancellation, we are left with the first term and a part of the second term, \( 1 + \frac{1}{2} \), giving us a clear convergence to 1.5. There are more complex convergence tests that can be employed for less straightforward series, but the understanding of partial fractions and series summation is imperative to tackle series convergence with confidence.
Applying the knowledge that successive terms cancel each other out, we recognize that the series converges to a finite number, because only a few terms of the series do not cancel and contribute to the final sum. Specifically, after cancellation, we are left with the first term and a part of the second term, \( 1 + \frac{1}{2} \), giving us a clear convergence to 1.5. There are more complex convergence tests that can be employed for less straightforward series, but the understanding of partial fractions and series summation is imperative to tackle series convergence with confidence.
Other exercises in this chapter
Problem 19
Approximate the sum of the series by using the first six terms. (See Example 4.) $$ \sum_{n=0}^{\infty} \frac{(-1)^{n} 2}{n !} $$
View solution Problem 19
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 20
Write the next two apparent terms of the sequence. Describe the pattern you used to find these terms. \(1,-\frac{3}{2}, \frac{9}{4},-\frac{27}{8}, \ldots\)
View solution Problem 20
Use the Integral Test to determine the convergence or divergence of the \(p\) -series. $$ \sum_{n=1}^{\infty} \frac{1}{n^{1 / 3}} $$
View solution