Problem 7
Question
Find an expression for the \(n\) th term of the sequence. (Assume that the pattern continues.) \(\left\\{\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \ldots\right\\}\)
Step-by-Step Solution
Verified Answer
The short answer for the nth term of the given sequence is \(a_n = \frac{n}{n+1}\).
1Step 1: Identify the pattern
Looking closely at the sequence, we can notice that the numerator of each term is 1 less than the denominator. For example, in the first term, the numerator is 1 and the denominator is 2, then for the second term, the numerator is 2 and the denominator is 3, so on and so forth.
2Step 2: Generate a general expression for the sequence
Now that we have observed a pattern in the way the sequence is formed, we can generate a general expression as a function of the term number, 'n'. Since the numerator is always 1 less than the denominator, the nth term will have a numerator of 'n' and a denominator of 'n+1', and hence the expression for the nth term of the sequence is as follows:
\[
a_n = \frac{n}{n+1}
\]
The given expression, \(a_n = \frac{n}{n+1}\), is the nth term of the sequence and can be used to find any term in the sequence by simply substituting the value of 'n'.
Key Concepts
Arithmetic SequencesPattern RecognitionGeneral Term Formula
Arithmetic Sequences
An arithmetic sequence is a list of numbers with a common difference between consecutive terms. This difference is constant across all the terms in the sequence. It allows us to predict and find any term when we know the previous terms.
In general, an arithmetic sequence can be expressed with the formula:
In our exercise, however, the sequence given is not arithmetic because the pattern does not include adding or subtracting a constant difference. Instead, it involves a relationship between the numerator and denominator of each fraction.
In general, an arithmetic sequence can be expressed with the formula:
- First term, denoted as \(a_1\), which is the starting point.
- Common difference, denoted as \(d\), which is added to each term to get the next one.
In our exercise, however, the sequence given is not arithmetic because the pattern does not include adding or subtracting a constant difference. Instead, it involves a relationship between the numerator and denominator of each fraction.
Pattern Recognition
Pattern recognition is a crucial skill in identifying and understanding sequences. It involves looking at the sequence and spotting trends or relationships between terms.
In our given sequence \[\left\{ \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \ldots \right\}\],
we observe a specific pattern:
In our given sequence \[\left\{ \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \ldots \right\}\],
we observe a specific pattern:
- Each numerator increases by 1, starting from 1, 2, 3, ..., which aligns with the term number \(n\).
- Each denominator is 1 more than its corresponding numerator.
General Term Formula
The general term formula is an equation that provides a way to calculate any term in a sequence without listing all preceding terms. It encapsulates the pattern of the sequence into a single expression.
For the given sequence \(\left\{\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \ldots\right\}\),
once we recognize the pattern, we can establish the general term formula as:
For the given sequence \(\left\{\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \ldots\right\}\),
once we recognize the pattern, we can establish the general term formula as:
- Denote the nth term as \(a_n\).
- Observe from the pattern that \(a_n = \frac{n}{n+1}\).
Other exercises in this chapter
Problem 7
Determine whether the series converges or diverges. $$ \sum_{n=2}^{\infty} \frac{(-1)^{n-1} \sqrt{n+1}}{n-1} $$
View solution Problem 7
Use the Integral Test to determine whether the series is convergent or divergent. $$ \sum_{n=1}^{\infty} \frac{n}{\left(n^{2}+1\right)^{3 / 2}} $$
View solution Problem 8
Determine whether the series is convergent, absolutely convergent, conditionally convergent, or divergent. \(\sum_{n=1}^{\infty} \frac{(-1)^{n-1} n}{\sqrt{2 n^{
View solution Problem 8
Use the Comparison Test to determine whether the series is convergent or divergent. \(\sum_{n=3}^{\infty} \frac{3^{n}}{2^{n}-4}\)
View solution