Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find an explicit formula a \(a_{n}=\) ____ for each sequence, determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\). $$ 1, \frac{2}{2^{2}-1^{2}}, \frac{3}{3^{2}-2^{2}}, \frac{4}{4^{2}-3^{2}}, \ldots $$

Step-by-Step Solution

Answer

The sequence is \(a_n = \frac{n}{2n-1}\) and it converges to \(\frac{1}{2}\).

Step 1: Identify the Pattern in the Sequence

The sequence is given as \(1, \frac{2}{2^{2}-1^{2}}, \frac{3}{3^{2}-2^{2}}, \frac{4}{4^{2}-3^{2}}, \ldots\). Notice that the numerator of each term corresponds to the index of the term, \(n\), in the sequence. The denominator is formulated as a difference of squares, specifically \(n^2 - (n-1)^2\).

Step 2: Simplify the Denominator

Write the denominator \(n^2 - (n-1)^2\) in its expanded form: \[n^2 - (n-1)^2 = n^2 - (n^2 - 2n + 1) = 2n - 1\]. This simplification shows that each term in the sequence can be expressed as \(\frac{n}{2n-1}\).

Step 3: Write the Explicit Formula

Based on the analysis in Step 1 and Step 2, the explicit formula that generates this sequence is \[a_n = \frac{n}{2n-1}\].

Step 4: Determine Convergence or Divergence

To find if the sequence converges, evaluate the limit \( \lim_{n \to \infty} a_n \) using the explicit formula:\[\lim_{n \to \infty} \frac{n}{2n-1} = \lim_{n \to \infty} \frac{n}{2n(1 - \frac{1}{2n})} = \lim_{n \to \infty} \frac{1}{2 - \frac{1}{n}}\].As \(n\) approaches infinity, \(\frac{1}{n}\) approaches 0, so the limit becomes \(\frac{1}{2}\).

Step 5: Conclusion on Convergence

The sequence \( a_n = \frac{n}{2n-1} \) converges to \( \frac{1}{2} \).

Key Concepts

Explicit Formula

In mathematics, an explicit formula is a clear and direct expression that specifies each term of a sequence. Without an explicit formula, calculating terms in a sequence would be cumbersome as we would have to rely on preceding terms to find the next one.
For sequences like the one we are examining, the goal is to find a formula that directly computes the nth term without needing any prior terms.
The sequence in our example is:

• 1
• \(\frac{2}{2^2 - 1^2}\)
• \(\frac{3}{3^2 - 2^2}\)
• \(\frac{4}{4^2 - 3^2}\)

By identifying patterns, we observe the numerator is always the term number \(n\), and the denominator is a difference of squares \(n^2 - (n-1)^2\), which simplifies to \(2n-1\).
Thus, the explicit formula becomes \(a_n = \frac{n}{2n-1}\).
This allows us to find any term in the sequence quickly and efficiently.

Limit of a Sequence

The limit of a sequence provides valuable information about the behavior of terms as they extend towards infinity. It helps us understand whether a sequence converges (settles to a single value) or diverges (does not settle to any value).
When we find the limit of a sequence, particularly an infinite sequence, we determine where the sequence "ends up" as the number of terms becomes very large. Analyzing sequences often requires evaluating this limiting behavior to understand convergence.
In the explicit formula \(a_n = \frac{n}{2n-1}\), to find the limit as \(n\) approaches infinity, we rearrange the expression:

• Divide numerator and denominator by \(n\) to express it as \( \frac{1}{2 - \frac{1}{n}} \).
• Recognize that as \(n\) becomes very large, \(\frac{1}{n}\) becomes very small, approaching zero.
• This simplifies to \(\frac{1}{2}\).

This result indicates that the sequence converges to \(\frac{1}{2}\). Hence, regardless of how large \(n\) gets, the terms of the sequence will come closer and closer to this value.

Difference of Squares

A difference of squares is a particular mathematical expression that takes the form \(a^2 - b^2\). This expression can be simplified using the identity \((a-b)(a+b) = a^2 - b^2\). This technique significantly simplifies expressions and aids in solving a variety of mathematical problems.
In dealing with sequences, this concept allows us to break down and simplify complex denominators when needed.
For our sequence, each term originally had a denominator of \((n^2 - (n-1)^2)\), where \(n\) was the term number. Using the identity for difference of squares, we simplified this to \(2n - 1\).
Simplifying complex algebraic expressions to simpler forms helps in both theoretical understanding and practical computation of sequences. By reducing \(n^2 - (n-1)^2\) to \(2n-1\), we made subsequent calculations to find the limit much more straightforward.