Problem 5
Question
Evaluate \(\lim _{n \rightarrow \infty} a_{n}\) for the given sequence \(\left\\{a_{n}\right\\}\). $$ a_{n}=\frac{3^{n}+2 \cdot 5^{n}}{2^{n}+3 \cdot 5^{n}} $$
Step-by-Step Solution
Verified Answer
The limit is \(\frac{2}{3}\).
1Step 1: 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.
2Step 2: Simplify the Sequence
To simplify the sequence, divide every term in the numerator and the denominator by the dominant term \(5^n\):\[a_n = \frac{\frac{3^n}{5^n} + \frac{2 \cdot 5^n}{5^n}}{\frac{2^n}{5^n} + \frac{3 \cdot 5^n}{5^n}} = \frac{\left(\frac{3}{5}\right)^n + 2}{\left(\frac{2}{5}\right)^n + 3}\]
3Step 3: 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:\[\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{0 + 2}{0 + 3} = \frac{2}{3}\]
4Step 4: 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.
Key Concepts
Dominant Term IdentificationSequence SimplificationExponential Growth Comparison
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.
Other exercises in this chapter
Problem 5
The given series may be shown to converge by using the Alternating Series Test. Show that the hypotheses of the Alternating Series Test are satisfied. $$ \sum_{
View solution Problem 5
Use the Comparison Test for Convergence to show that the given series converges. State the series that you use for comparison and the reason for its convergence
View solution Problem 5
State what conclusion, if any, may be drawn from the Divergence Test. $$ \sum_{n=1}^{\infty} \frac{1}{1+1 / n} $$
View solution Problem 6
Express the given function as a power series in \(x\) with base point \(0 .\) Calculate the radius of convergence \(R\). \(\frac{1}{9-x^{2}}\)
View solution