Problem 4 - Calculus Single Variable

Question

Evaluate \(\lim _{n \rightarrow \infty} a_{n}\) for the given sequence \(\left\\{a_{n}\right\\}\). $$ a_{n}=\frac{3^{2 n}+2}{9^{n+1}+1} $$

Step-by-Step Solution

Answer

The limit of the sequence is \(\frac{1}{9}.\)

Step 1: Express Components in Terms of Base 3

Observe that the sequence is given by \[ a_{n} = \frac{3^{2n} + 2}{9^{n+1} + 1}. \] First, rewrite everything in the fraction in terms of powers of 3. Notice that \( 9^{n+1} \) can be written as \( (3^2)^{n+1} = 3^{2(n+1)}\). Thus, we have the sequence as \[ a_{n} = \frac{3^{2n} + 2}{3^{2n+2} + 1}. \]

Step 2: Simplify the Fraction

Divide both the numerator and the denominator by the highest power of 3 present in the denominator, which is \(3^{2n+2}\). This gives us: \[ a_n = \frac{\frac{3^{2n}}{3^{2n+2}} + \frac{2}{3^{2n+2}}}{\frac{3^{2n+2}}{3^{2n+2}} + \frac{1}{3^{2n+2}}}.\]Simplifying further, we have: \[ a_n = \frac{\frac{1}{3^2} + \frac{2}{3^{2n+2}}}{1 + \frac{1}{3^{2n+2}}} \] which simplifies to \[ a_n = \frac{\frac{1}{9} + \frac{2}{3^{2n+2}}}{1 + \frac{1}{3^{2n+2}}}. \]

Step 3: Evaluate as \(n\) Approaches Infinity

As \(n\) approaches infinity, \(\frac{2}{3^{2n+2}} \) and \(\frac{1}{3^{2n+2}} \) approach zero, as the denominators grow much faster than the numerators. Therefore, we can approximate: \[ \lim_{n \to \infty} a_n \approx \frac{\frac{1}{9}}{1} = \frac{1}{9}. \]

Step 4: Conclusion

After removing the components that approach zero as \(n\) becomes very large, the limiting behavior of the sequence \(a_n\) simplifies to constant value. Therefore, we conclude: \[ \lim_{n \rightarrow \infty} a_{n} = \frac{1}{9}. \]

Key Concepts

Infinite Sequences

Infinite sequences are an important concept in mathematics, describing a list of numbers in a specific order that extends indefinitely. In simpler terms, an infinite sequence does not stop; it goes on forever.
When dealing with sequences, we often use a function rule to determine how each term is calculated. For example, the sequence in our exercise uses the formula \( a_n = \frac{3^{2n} + 2}{9^{n+1} + 1} \). Here, \( n \) represents the position of the term in the sequence.
One key aspect of infinite sequences is their behavior as \( n \) becomes very large. Mathematicians study this to understand whether the sequence "settles down" as it progresses. This investigation is what's known as finding the **limit** of a sequence.

Limits and Continuity

Limits and continuity help us understand how functions and sequences behave as the input values grow larger. Think of limits like the values that a function or sequence approaches as we move towards infinity.
For sequences, the limit is the number the terms get closer to as \( n \) increases. In our problem, we determined that the limit of the sequence \( a_n = \frac{3^{2n} + 2}{9^{n+1} + 1} \) is \( \frac{1}{9} \). This means that as \( n \) grows without bound, the terms of the sequence get very close to \( \frac{1}{9} \).
To find limits of sequences like this one, we often simplify the sequence by dividing by the highest power of the growing factor, typically found in the denominator. This allows us to focus on significant terms while dismissing smaller, less relevant components as they shrink near zero.

Convergence of Sequences

Convergence describes whether the terms of a sequence get close to a certain number -- its limit -- as \( n \) increases. A sequence is said to "converge" if it approaches a particular value as \( n \) goes to infinity.
In simple terms, convergence is akin to achieving stability in values over time. For our sequence \( a_n \), convergence is seen by calculating the limit—and confirming it approaches \( \frac{1}{9} \).
To verify convergence:

• We simplify the sequence, as shown by rewriting in terms of the largest power present.
• We then check that smaller components diminish as \( n \) becomes very large.

Ultimately, the sequence's behavior at infinity gives insight into convergence, helping us understand long-term trends and outcomes in mathematics.