Problem 6
Question
Determine whether the given sequence converges or diverges. \(\left\\{\frac{n i+2^{n}}{3 n i+5^{n}}\right\\}\)
Step-by-Step Solution
Verified Answer
The sequence converges to 0.
1Step 1: Understanding the Sequence
The sequence is given by \( a_n = \frac{n i + 2^n}{3 n i + 5^n} \). We are asked to determine whether this sequence converges or diverges.
2Step 2: Identifying Leading Terms
To analyze the behavior of the sequence, identify the leading terms in the numerator and the denominator. In this case, \(2^n\) is the leading term in the numerator and \(5^n\) in the denominator as \(n\) becomes very large.
3Step 3: Simplifying the Sequence
Divide both the numerator and the denominator by \(5^n\), the leading term in the denominator. The sequence becomes \( a_n = \frac{\frac{n i}{5^n} + \left(\frac{2}{5}\right)^n}{\frac{3 n i}{5^n} + 1} \).
4Step 4: Analyzing the Limit
As \(n\) approaches infinity, \(\frac{n i}{5^n} \to 0\) and \(\left(\frac{2}{5}\right)^n \to 0\), because \(\frac{2}{5} < 1\). Therefore, the sequence simplifies to \( a_n = \frac{0 + 0}{0 + 1} = 0 \).
5Step 5: Conclusion
Since the sequence \( a_n = 0 \) for large values of \(n\), the sequence converges to \(0\).
Key Concepts
LimitsExponential FunctionsComplex Numbers
Limits
Understanding the concept of limits is crucial when determining the behavior of sequences as they extend towards infinity. Essentially, a limit analyzes what value a sequence or function approaches as the input grows arbitrarily large. In mathematical notation, if becomes very large (approaches infinity), and the output of the function approaches a particular number, we signify this with:
When working with sequences, observing the limits of sequences involving exponential terms is vital since exponential functions tend to dominate the sequence's behavior. In our exercise, we applied the concept of limits to investigate the behavior of complex components and base fractions, specifically resolving whether terms approach zero.
This sequence indeed converges to zero primarily because each component in the sequence approaches zero faster than any linear or polynomial term. The analysis concludes that regardless of the complex number \(i\), the overall outcome is dominated by the exponential terms.
- \( \lim_{{n \to \infty}} a_n = L \)
When working with sequences, observing the limits of sequences involving exponential terms is vital since exponential functions tend to dominate the sequence's behavior. In our exercise, we applied the concept of limits to investigate the behavior of complex components and base fractions, specifically resolving whether terms approach zero.
This sequence indeed converges to zero primarily because each component in the sequence approaches zero faster than any linear or polynomial term. The analysis concludes that regardless of the complex number \(i\), the overall outcome is dominated by the exponential terms.
Exponential Functions
Exponential functions play a significant role in the study of sequence convergence. They are functions of the form \( a^n \), where \(a\) is a constant only when \(a\) is greater than 1, and particularly influential when compared to linear or fractional expressions within sequences.
Their growth or decay significantly impacts a sequence’s behavior, often overriding linear terms. In our given sequence:
This allowed the dominant exponential terms to clarify whether the sequence converged or diverged. Recognizing how exponential functions grow or decay faster clarifies why sequences like these often trend towards specific limits.
Their growth or decay significantly impacts a sequence’s behavior, often overriding linear terms. In our given sequence:
- The term \(2^n\) in the numerator became negligible compared to \(5^n\) in the denominator because \( \left(\frac{2}{5}\right)^n \) reduces towards zero as \(n\) increases since \(\frac{2}{5} < 1\).
This allowed the dominant exponential terms to clarify whether the sequence converged or diverged. Recognizing how exponential functions grow or decay faster clarifies why sequences like these often trend towards specific limits.
Complex Numbers
Complex numbers, composed of a real part and an imaginary part, are notable for their appearance in various mathematical contexts including sequence calculations. In our scenario, the imaginary unit \(i\) has been used within the sequence:
Consider \(\lim_{{n \to \infty}} \frac{n i}{5^n} = 0\). The imaginary part alone cannot dominate over an exponential decline, such as \((\frac{2}{5})^n\), which approaches zero. Therefore, even with the presence of complex numbers in this sequence, the limit conclusion remains unaffected, leading us to determine sequence convergence effectively.
- \(n i+2^n\)
- \(3 n i+5^n\)
Consider \(\lim_{{n \to \infty}} \frac{n i}{5^n} = 0\). The imaginary part alone cannot dominate over an exponential decline, such as \((\frac{2}{5})^n\), which approaches zero. Therefore, even with the presence of complex numbers in this sequence, the limit conclusion remains unaffected, leading us to determine sequence convergence effectively.
Other exercises in this chapter
Problem 5
In Problems 5-10, determine whether the given sequence converges or diverges. $$ \left\\{\frac{3 n i+2}{n+n i}\right\\} $$
View solution Problem 6
Expand the given function in a Maclaurin series. Give the radius of convergence of each series. \(f(z)=z e^{-z^{2}}\)
View solution Problem 6
Evaluate the given trigonometric integral. \(\int_{0}^{\pi} \frac{1}{1+\sin ^{2} \theta} d \theta\)
View solution Problem 6
Use a Laurent series to find the indicated residue. \(f(z)=\frac{e^{-z}}{(z-2)^{2}} ;\) Res \((f(z), 2)\)
View solution