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Chapter 8: Problem 5

Evaluate \(\lim _{n \rightarrow \infty} a_{n}\) for the given sequence \(\left\\{a_{n}\right\\}\). $$ a_{n}=\frac{3^{n}+2 \cdot 5^{n}}{2^{n}+3 \cdot 5^{n}} $$

Short Answer

Expert verified
The limit is \(\frac{2}{3}\).

Step by step solution

01

Identify Dominant Terms

In the sequence \(a_n = \frac{3^n + 2 \cdot 5^n}{2^n + 3 \cdot 5^n}\), the dominant term in both the numerator and the denominator is \(5^n\). This is because \(5^n\) grows faster than both \(3^n\) and \(2^n\) as \(n\) approaches infinity.
02

Simplify the Sequence

To simplify the sequence, divide every term in the numerator and the denominator by the dominant term \(5^n\):\[a_n = \frac{\frac{3^n}{5^n} + \frac{2 \cdot 5^n}{5^n}}{\frac{2^n}{5^n} + \frac{3 \cdot 5^n}{5^n}} = \frac{\left(\frac{3}{5}\right)^n + 2}{\left(\frac{2}{5}\right)^n + 3}\]
03

Evaluate the Limit

As \(n\) approaches infinity, the terms \(\left(\frac{3}{5}\right)^n\) and \(\left(\frac{2}{5}\right)^n\) both approach 0 because they are less than 1 and raised to an increasing power. Thus, the sequence simplifies to:\[\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{0 + 2}{0 + 3} = \frac{2}{3}\]
04

Conclusion

The limit of the sequence has been evaluated based on the simplifications made. The terms \(\left(\frac{3}{5}\right)^n\) and \(\left(\frac{2}{5}\right)^n\) vanish as they approach zero when \(n\) is very large.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dominant Term Identification
When working with sequences and looking to find limits, a crucial step involves identifying the dominant term. This requires recognizing which parts of a sequence grow faster or more significantly as the sequence progresses.* For the sequence given, \(a_n = \frac{3^n + 2 \cdot 5^n}{2^n + 3 \cdot 5^n}\), we see multiple terms in both the numerator and denominator.* The terms with exponential parts, like \(5^n\), are especially important since exponential growth can become very large very fast as \(n\) increases.* Among \(3^n\), \(2 \cdot 5^n\), and \(2^n\), \(3 \cdot 5^n\), the term \(5^n\) in both the numerator and denominator is identified as dominant.* Recognizing \(5^n\) as the dominant term is key because it outpaces the growth of \(3^n\) and \(2^n\) significantly, particularly as \(n\) moves toward infinity.
Sequence Simplification
After identifying the dominant term, you can often simplify complex expressions by 'canceling out' the dominant parts. In this case, that's achieved by dividing both the numerator and the denominator by \(5^n\).* The process uses the fact that when you divide any of these terms by the dominant term (\(5^n\)), the ones involving \(5^n\) simplify nicely, becoming whole numbers or terms that shrink.* This gives us the new form: \(a_n = \frac{\left(\frac{3}{5}\right)^n + 2}{\left(\frac{2}{5}\right)^n + 3}\).* Here, realistically, the same reasoning for dominant terms applies: as \(n\) grows larger, fractions involving powers of numbers less than one approach zero.* Simplifying in this way reduces the complexity, making it easier to evaluate limits and predict sequence behavior at infinity.
Exponential Growth Comparison
Understanding the growth rates of exponential terms is vital in limit evaluation of sequences. Let's inspect the sequence elements involving exponential growth:* Each base, \(3\), \(2\), and \(5\), raised to a power \(n\) exhibits different growth rates.* In exponential terms, larger bases mean faster growth. Therefore, terms like \(5^n\) expand more swiftly than \(3^n\) or \(2^n\).* For this reason, in very large \(n\), terms \(\left(\frac{3}{5}\right)^n\) and \(\left(\frac{2}{5}\right)^n\) converge to zero due to their base being less than one.* It's crucial to compare these growth rates logically, ensuring sequence simplification is effective. This fundamentally changes a sequence's nature when calculating limits, as slower growing or shrinking terms become negligible.

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Most popular questions from this chapter

Let \(\left\\{a_{n}\right\\}\) be a sequence of positive numbers. In a course on mathematical analysis, one learns that if the two limits \(\lim _{n \rightarrow \infty} a_{n+1} / a_{n}\) and \(\lim _{n \rightarrow \infty} a_{n}^{1 / n}\) exist, then they are equal. In each of Exercises \(65-68\), produce a plot that illustrates the equality of these two limits. Your plot should include a horizontal line that is the asymptote of the points \(\left\\{\left(n, a_{n+1} / a_{n}\right)\right\\}\) and \(\left\\{\left(n, a_{n}^{1 / n}\right)\right\\}\). \(a_{n}=e^{n} / \ln (1+n)\)

In each of Exercises 91-94 a function \(f\), a base point \(c\), and a point \(x_{0}\) are given. Plot \(y=\left|f^{(3)}(t)\right|\) for \(t\) between \(c\) and \(x_{0}\). Use your plot to estimate the quantity \(M\) of Theorem 2 . Then use your value of \(M\) to obtain an upper bound for the absolute error \(\left|R_{2}\left(x_{0}\right)\right|=\left|f\left(x_{0}\right)-T_{2}\left(x_{0}\right)\right|\) that results when \(f\left(x_{0}\right)\) is approximated by the order 2 Taylor polynomial with base point \(c\). $$ f(x)=\ln \left(2 x^{2}-1\right) \quad c=1 \quad x_{0}=1.3 $$

In each of Exercises 91-94 a function \(f\), a base point \(c\), and a point \(x_{0}\) are given. Plot \(y=\left|f^{(3)}(t)\right|\) for \(t\) between \(c\) and \(x_{0}\). Use your plot to estimate the quantity \(M\) of Theorem 2 . Then use your value of \(M\) to obtain an upper bound for the absolute error \(\left|R_{2}\left(x_{0}\right)\right|=\left|f\left(x_{0}\right)-T_{2}\left(x_{0}\right)\right|\) that results when \(f\left(x_{0}\right)\) is approximated by the order 2 Taylor polynomial with base point \(c\). $$ f(x)=\sqrt{16+x^{2}} \quad c=3 \quad x_{0}=2.4 $$

Use a Taylor polynomial with base point \(c=e^{3}\) to approximate \(\ln (20)\) to five decimal places. Do not use a calculator to evaluate any value of \(\ln (x),\) but you may use a calculator for arithmetic with the number \(e\) and its powers.

Suppose that \(r\) is a constant greater than \(1 .\) Calculate the \(N^{\text {th }}\) partial sum of \(\sum_{n=1}^{\infty} \frac{r^{n}}{\left(r^{n}-1\right)\left(r^{n+1}-1\right)},\) and use the formula to evaluate the infinite series.
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