Question

Evaluate the limit of the following sequences. $$a_{n}=\frac{75^{n-1}}{99^{n}}+\frac{5^{n} \sin n}{8^{n}}$$

Short answer

Answer: The limit of the sequence \(a_n\) as \(n\) approaches infinity is 0.

Step-by-step solution

Simplify the expression

We can rewrite the equation as: $$a_n = \frac{75^0}{99} \cdot \frac{75^{n-1}}{99^{n-1}} + \frac{5^n \sin n}{8^n}$$ Which becomes: $$a_n = \frac{1}{99} \cdot \left(\frac{75}{99}\right)^{n-1} + \left(\frac{5}{8}\right)^n \sin n$$ Now, we can analyze each term as \(n\) approaches infinity.

Find the limit of the first term

We want to find the limit as \(n\) approaches infinity for the first term: $$\lim_{n \to \infty} \frac{1}{99} \cdot \left(\frac{75}{99}\right)^{n-1}$$ Since \(0 < \frac{75}{99} < 1\), and for any \(r\) where \(0 < r < 1\), \(\lim_{n \to \infty} r^n = 0\), we can simplify this to: $$\lim_{n \to \infty} \frac{1}{99} \cdot 0 = 0$$

Find the limit of the second term

We want to find the limit as \(n\) approaches infinity for the second term: $$\lim_{n \to \infty} \left(\frac{5}{8}\right)^n \sin n$$ The sine function varies between -1 and 1, so we want to find \(\lim_{n \to \infty} (\frac{5}{8})^n\) and then utilize the squeeze theorem. Since \(0 < \frac{5}{8} < 1\), we can simplify this to: $$\lim_{n \to \infty} \left(\frac{5}{8}\right)^n = 0$$ Now, using the squeeze theorem, we know that: $$-1 \le \sin n \le 1$$ Multiplying the inequality by \(\left(\frac{5}{8}\right)^n\): $$0 \le \left(\frac{5}{8}\right)^n \le \left(\frac{5}{8}\right)^n \sin n$$ Since \(\lim_{n \to \infty} \left(\frac{5}{8}\right)^n = 0\), this means the limit of the second term is also 0.

Find the limit of the entire sequence

Now that we have the limits of each term within the sequence, we can find the entire sequence's limit as \(n\) approaches infinity: $$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \left[\frac{1}{99} \cdot \left(\frac{75}{99}\right)^{n-1} + \left(\frac{5}{8}\right)^n \sin n\right] $$ Using our previous results, we can plug in the limits of each term: $$\lim_{n \to \infty} a_n = 0 + 0 = 0$$ Therefore, the limit of the sequence \(a_n\) as \(n\) approaches infinity is 0.

Key concepts

Squeeze Theorem

The Squeeze Theorem is an important concept in mathematics, particularly useful for finding the limits of sequences and functions. When we deal with a sequence or function that is difficult to evaluate directly, the Squeeze Theorem can help us simplify the problem by bounding it between two more manageable sequences or functions.
The idea is simple:

• Suppose you have three sequences or functions: \(a_n\), \(b_n\), and \(c_n\).
• Let's say \(a_n \le b_n \le c_n\) for all values of \(n\) or for all values in the domain.
• If both \(\lim_{n \to \infty} a_n = L\) and \(\lim_{n \to \infty} c_n = L\), then by the Squeeze Theorem, \(\lim_{n \to \infty} b_n = L\) as well.

This theorem is particularly useful when dealing with oscillatory terms, such as sine or cosine functions, which are bound between -1 and 1. In our exercise, we apply the Squeeze Theorem to the term \(\left(\frac{5}{8}\right)^n \sin n\). Since \(\sin n\) is always between -1 and 1, multiplying it by \(\left(\frac{5}{8}\right)^n\), which becomes smaller as \(n\) becomes larger, heads towards zero, confirming that the entire term approaches a limit of zero.

Convergence of Sequences

Understanding the convergence of sequences is crucial in calculus and analysis. Convergence happens when a sequence approaches a specific value as \(n\) (the sequence term index) goes to infinity. Not every sequence will converge; some may diverge or oscillate indefinitely.
In determining convergence:

• A sequence \(a_n\) is said to converge to a limit \(L\) if, for any small positive number \(\epsilon\), there exists a sufficiently large \(N\) such that for all \(n > N\), the terms satisfy \(|a_n - L| < \epsilon\).
• This means that as you progress in the sequence, the terms eventually approximate \(L\) as closely as desired.

In our exercise, we evaluated the limit of the sequence \(a_n\) as \(n\) approaches infinity. Both components of the sequence \(\left(\frac{75}{99}\right)^{n-1}\) and \(\left(\frac{5}{8}\right)^n\) were shown to individually approach zero, hence driving the entire sequence to converge to zero. This process is essential in understanding how sequences behave over prolonged periods and often forms the foundation for advanced mathematical analysis.

Exponential Decay

Exponential decay describes a process where each term in a sequence or function becomes progressively smaller by a fixed percentage, often expressed as a factor per step or unit of time. This concept is seen widely in both mathematics and real-world applications such as radioactive decay, population decline, and capacitor discharge in electronics.
More formally, a sequence \(a_n = r^n\) where \(0 < r < 1\) is characterized by exponential decay because \(r^n\) becomes smaller as \(n\) increases.

• The smaller \(r\) is compared to 1, the faster the sequence approaches zero.
• Such sequences are critical in many fields, indicating processes that eventually dwindle to negligible amounts.

In our specific problem, both \(\left(\frac{75}{99}\right)^{n-1}\) and \(\left(\frac{5}{8}\right)^n\) demonstrate exponential decay. With the base values \(\frac{75}{99}\) and \(\frac{5}{8}\) being less than 1, each increment in \(n\) brings us closer to zero, making the limit evaluations straightforward. This decay is why both terms approach zero, illustrating how exponential decay influences the convergence of sequences.