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Converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series’ convergence or divergence.) $$ \sum_{n=1}^{\infty} \frac{5^{n}}{4^{n}+3} $$

Short Answer

Expert verified
The series diverges because \( \left( \frac{5}{4} \right)^n \) diverges.

Step by step solution

01

Analyze the structure of the series

The series we have is an infinite series of the form \( \sum_{n=1}^{\infty} \frac{5^{n}}{4^{n}+3} \). Notice the numerator is an exponential function \( 5^n \) and the denominator is \( 4^n + 3 \). We can analyze the behavior of the terms as \( n \to \infty \).
02

Simplify each term for large n

As \( n \to \infty \), the term \( \frac{5^{n}}{4^{n}+3} \) behaves like \( \frac{5^n}{4^n} \) since \( 4^n \) grows much faster than \( 3 \). This simplifies to \( \left( \frac{5}{4} \right)^n \).
03

Apply the Geometric Series Test

The terms can be approximated by a geometric sequence \( \left( \frac{5}{4} \right)^n \). A geometric series of the form \( \sum_{n=1}^{\infty} a r^n \) converges if \( |r| < 1 \) and diverges if \( |r| \geq 1 \). Here, \( r = \frac{5}{4} > 1 \), so the series diverges.
04

Conclusion

Since the series \( \sum_{n=1}^{\infty} \frac{5^{n}}{4^{n}+3} \) can be compared to a divergent geometric series \( \left( \frac{5}{4} \right)^n \) and \( \frac{5}{4} > 1 \), this series also diverges.

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Key Concepts

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

Geometric Series Test
The Geometric Series Test is a powerful tool for determining the convergence or divergence of an infinite series. This applies particularly to geometric series of the form \( \sum_{n=1}^{\infty} ar^n \). In this context, \( a \) is the first term, and \( r \) is the common ratio of the series.
The main rule for this test is simple:
  • If the absolute value of the common ratio \( |r| \) is less than 1, the series converges.
  • If \( |r| \) is greater than or equal to 1, the series diverges.
In the given problem, the series could be approximated as a geometric series \( \left( \frac{5}{4} \right)^n \) for large \( n \). Here, the common ratio \( r = \frac{5}{4} \) is greater than 1, indicating divergence according to the Geometric Series Test. This makes it clear and easy to understand the behavior of the series.
Exponential Functions
Exponential functions are functions of the form \( a^n \), where \( a \) is a constant and \( n \) represents the exponent. These functions are characterized by their rapid growth or decay, depending on the value of \( a \).
- If \( a > 1 \), the function grows exponentially.- If \( 0 < a < 1 \), the function decays exponentially.In the series \( \sum_{n=1}^{\infty} \frac{5^{n}}{4^{n}+3} \), the numerator \( 5^n \) is an exponential function that grows steadily as \( n \) increases. Comparatively, the term \( 4^n + 3 \) in the denominator can be approximated by \( 4^n \) for large \( n \), since \( 4^n \) dominates \( 3 \).
Understanding the behavior of exponential functions helps in breaking down complex series terms, simplifying comparison with geometric sequences, and applying convergence tests like the Geometric Series Test.
Divergence Analysis
When determining the convergence or divergence of a series, divergence analysis involves examining whether the series grows unbounded. We typically check the limit of the series terms as \( n \to \infty \).
In the given problem, the focus is on terms \( \frac{5^n}{4^n + 3} \) for large \( n \). As \( n \to \infty \), the term \( 5^n \) in the numerator grows faster than \( 4^n \). Evaluating the limit turns the fraction into \( \left( \frac{5}{4} \right)^n \), demonstrating an increase with each term.
The divergence analysis reveals that \( \left( \frac{5}{4} \right)^n \) essentially becomes a base for comparison, guiding us toward a conclusion that the sequence does not converge towards zero. Instead, because the limit of the terms does not equal zero, and the ratio is greater than 1, the series diverges. This analysis plays a crucial role, ensuring that such series behaviors are predicted accurately.

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Most popular questions from this chapter

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Compound interest, deposits, and withdrawals If you invest an amount of money \(A_{0}\) at a fixed annual interest rate \(r\) compounded \(m\) times per year, and if the constant amount \(b\) is added to the account at the end of each compounding period (or taken from the account if \(b<0 ),\) then the amount you have after \(n+1\) compounding periods is $$ A_{n+1}=\left(1+\frac{r}{m}\right) A_{n}+b $$ a. If \(A_{0}=1000, r=0.02015, m=12,\) and \(b=50\) , calculate and plot the first 100 points \(\left(n, A_{n}\right) .\) How much money is in your account at the end of 5 years? Does \(\left\\{A_{n}\right\\}\) converge? Is \(\left\\{A_{n}\right\\}\) bounded? b. Repeat part (a) with \(A_{0}=5000, r=0.0589, m=12,\) and \(b=-50 .\) c. If you invest 5000 dollars in a certificate of deposit (CD) that pays 4.5\(\%\) annually, compounded quarterly, and you make no further investments in the CD, approximately how many years will it take before you have \(20,000\) dollars? What if the CD earns 6.25\(\% ?\) d. It can be shown that for any \(k \geq 0\) , the sequence defined recursively by Equation \((1)\) satisfies the relation $$ A_{k}=\left(1+\frac{r}{m}\right)^{k}\left(A_{0}+\frac{m b}{r}\right)-\frac{m b}{r} $$ For the values of the constants \(A_{0}, r, m,\) and \(b\) given in part (a), validate this assertion by comparing the values of the first 50 terms of both sequences. Then show by direct substitution that the terms in Equation \((2)\) satisfy the recursion formula in Equation ( 1\()\) .

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