Problem 67
Question
If the \( n \)th partial sum of a series \( \sum_{n = 1}^{\infty} a_n \) is \( s_n = \frac {n - 1}{n + 1} \) find \( a_n \) and \( \sum_{n = 1}^{\infty} a_n. \)
Step-by-Step Solution
Verified Answer
The series term is \( a_n = \frac{1}{n(n+1)} \), and the series sums to 1.
1Step 1: Understanding Partial Sums
The given sequence of partial sums is: \( s_n = \frac{n - 1}{n + 1} \). To find the series, we need to determine the general term \( a_n \) from this expression.
2Step 2: Determining the General Series Term \( a_n \)
Recall that the series term \( a_n \) can be found from the partial sums by the relation \( a_n = s_n - s_{n-1} \). Write the expression for \( s_{n-1} \): \( s_{n-1} = \frac{n-2}{n} \). Now subtract \( s_{n-1} \) from \( s_n \):\[a_n = \frac{n-1}{n+1} - \frac{n-2}{n} = \frac{n^2 - n - 2(n+1) + (n-2)}{n(n+1)} = \frac{1}{n(n+1)}\]
3Step 3: Recognizing the Series Pattern
Notice \( a_n = \frac{1}{n(n+1)} \) is a telescoping series. This is critical because a telescoping series can easily be summed as terms cancel out.
4Step 4: Summing the Series
Recognizing the telescoping nature, rewrite each term: \( a_n = \frac{1}{n} - \frac{1}{n+1} \). Thus, the series \( \sum_{n = 1}^{\infty} a_n \) simplifies substantially as terms cancel each other out in the series expansion.
5Step 5: Finding the Total Sum
Due to the telescoping property, when you write out the first few terms, you observe:\[\left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \cdots + \left( \frac{1}{n} - \frac{1}{n+1} \right) = 1\]Thus, the infinite series \( \sum_{n=1}^{\infty} a_n \) converges to 1.
Key Concepts
Telescoping SeriesGeneral TermInfinite SeriesSequence of Partial Sums
Telescoping Series
An important concept in calculus is the telescoping series. This type of series has terms that cancel each other out when summed. It simplifies the process of finding the sum of an infinite series. A telescoping series looks something like this:
- Each term can be expressed in a form that allows consecutive terms to cancel each other out when added.
- For example, in our series, we rewrote each term as \( a_n = \frac{1}{n} - \frac{1}{n+1} \).
- When summed, many inner terms cancel out, leaving only the first part of the first term and the second part of the last term.
General Term
In any sequence or series, identifying the general term \(a_n\) is crucial for understanding its behavior. The general term gives us a formula describing each term of the series in relation to 'n'. Here's how you can find it:
- From partial sums, the series term \(a_n\) can be derived by subtraction: \(a_n = s_n - s_{n-1}\).
- In our example, given \(s_n = \frac{n-1}{n+1}\), we found \(s_{n-1} = \frac{n-2}{n}\).
- Subtracting these yields \(a_n = \frac{1}{n(n+1)}\).
Infinite Series
An infinite series sums an infinite number of terms, providing a foundational element in higher mathematics. Specifically, it’s about evaluating the behavior of sums that extend indefinitely. Here's how we make sense of these seemingly endless series:
- Think of it as the limit of partial sums \( s_n \) as \( n \to \infty \).
- Unlike finite series, infinite series do not necessarily have a finite sum.
- In our example, the infinite series \( \sum_{n=1}^{\infty} a_n \) has been found to converge to a specific value, 1.
Sequence of Partial Sums
The sequence of partial sums \( s_n \) is the accumulation of the first 'n' terms of a series. It's instrumental in determining the properties of a series, especially its convergence:
- Any series has a corresponding sequence of partial sums, helping us trace the series' growth.
- We observe whether the sequence converges to a particular number.
- For example, given \( s_n = \frac{n-1}{n+1} \), we can see how each partial sum approaches a value.
- In our case, as \( n \to \infty \), \( s_n \to 1 \), thus pointing to the convergence of our infinite series to 1.
Other exercises in this chapter
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