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Chapter 11: Problem 20

Test the series for convergence or divergence. $$\sum_{k=1}^{\infty} \frac{k+5}{5^{k}}$$

Short Answer

Expert verified
The series converges by the Ratio Test.

Step by step solution

01

Identify the Series Type

The given series is \( \sum_{k=1}^{\infty} \frac{k+5}{5^k} \). This is a series with terms of the form \( \frac{k+5}{5^k} \). Notice that it combines a linear polynomial \( k+5 \) in the numerator with an exponential function \( 5^k \) in the denominator, suggesting a comparison with a geometric series or fitting for the ratio test.
02

Apply the Ratio Test

The ratio test evaluates the limit \( L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \). For our series, \( a_k = \frac{k+5}{5^k} \) and \( a_{k+1} = \frac{(k+1)+5}{5^{k+1}} = \frac{k+6}{5^{k+1}} \). Let's find this limit:\[L = \lim_{k \to \infty} \left| \frac{\frac{k+6}{5^{k+1}}}{\frac{k+5}{5^k}} \right| = \lim_{k \to \infty} \frac{(k+6)}{5(k+5)} \].
03

Simplify and Evaluate the Limit

Simplify the limit expression:\[L = \lim_{k \to \infty} \frac{k+6}{5(k+5)} = \frac{1}{5} \lim_{k \to \infty} \frac{k+6}{k+5} \].Now, divide both the numerator and the denominator by \( k \):\[L = \frac{1}{5} \times \lim_{k \to \infty} \frac{1 + \frac{6}{k}}{1 + \frac{5}{k}} \].As \( k \to \infty \), both \( \frac{6}{k} \) and \( \frac{5}{k} \) approach 0, so the limit becomes:\[L = \frac{1}{5} \times \frac{1}{1} = \frac{1}{5}.\]
04

Conclude Convergence or Divergence

According to the ratio test, if \( L < 1 \), the series converges. Since \( L = \frac{1}{5} < 1 \), the series \( \sum_{k=1}^{\infty} \frac{k+5}{5^k} \) converges.

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Key Concepts

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

Ratio Test
One of the methods to determine if a series converges or diverges is the Ratio Test. The Ratio Test involves checking the limit of the ratio of consecutive terms in the series.
  • The test examines the limit, denoted as \( L \), of \( \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \), where \( a_k \) is the general term of the series.
  • If \( L < 1 \), the series converges absolutely.
  • If \( L > 1 \) or \( L = \infty \), the series diverges.
  • If \( L = 1 \), the test is inconclusive, and other methods must be used.
In this exercise, we used the Ratio Test, with \( L = \frac{1}{5} \), which is less than 1. Therefore, it confirms that the series \( \sum_{k=1}^{\infty} \frac{k+5}{5^k} \) converges.
Geometric Series
A geometric series is a series with a constant ratio between successive terms. A typical geometric series takes the form \( a + ar + ar^2 + ar^3 + \ldots \), where \( a \) is the first term and \( r \) is the common ratio.
  • If the absolute value of the common ratio \( |r| < 1 \), the series converges. The sum can be calculated using the formula \( S = \frac{a}{1-r} \).
  • If \( |r| \geq 1 \), the series diverges.
In the solution, although \( \sum_{k=1}^{\infty} \frac{k+5}{5^k} \) isn't strictly a geometric series, the behavior of its terms primarily governed by the exponential denominator \( 5^k \) allows us to draw parallels with a geometric series, where the terms decrease exponentially.
Exponential Function Evaluation
Exponential functions are crucial in analyzing series as they can significantly affect convergence. In this exercise, we evaluated the series \( \sum_{k=1}^{\infty} \frac{k+5}{5^k} \), where the term \( 5^k \) appears in the denominator.
  • An exponential term like \( 5^k \) grows rapidly as \( k \) increases, generally forcing the term \( \frac{k+5}{5^k} \) towards zero, favoring convergence.
  • Even when the numerator \( k+5 \) grows linearly, its effect is minimized by the faster increase of \( 5^k \) in the denominator.
This interplay between the linear and exponential components is key in determining the series behavior and confirming convergence.
Limit Evaluation
Limit evaluation is a fundamental process in calculus and is crucial in the Ratio Test and similar tests. It involves simplifying expressions to understand behavior as \( k \) approaches infinity.
  • In the context of series convergence, limits help predict how series terms behave as they approach infinity.
  • For the ratio \( \frac{k+6}{5(k+5)} \), multiplying out the factors shows the eventual ratio depends largely on the leading terms.
  • By dividing both the numerator and the denominator by \( k \), we simplify the expression, revealing how smaller terms become negligible as \( k \) goes to infinity, leading to the conclusion that the limit evaluates to a known ratio.
This limit evaluation was performed step-by-step to achieve \( L = \frac{1}{5} \), solidifying our convergence assessment based on the Ratio Test.

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Most popular questions from this chapter

(a) Show that if \(\lim _{n \rightarrow \infty} a_{2 n}=L\) and \(\lim _{n \rightarrow \infty} a_{2 n+1}=L\) then \(\left\\{a_{n}\right\\}\) is convergent and \(\lim _{n \rightarrow \infty} a_{n}=L\) (b) If \(a_{1}=1\) and $$a_{n+1}=1+\frac{1}{1+a_{n}}$$ find the first eight terms of the sequence \(\left\\{a_{n}\right\\} .\) Then use part (a) to show that lim \(_{n \rightarrow \infty} a_{n}=\sqrt{2} .\) This gives the continued fraction expansion $$ \sqrt{2}=1+\frac{1}{2+\frac{1}{2+\cdots}} $$

The resistivity \(\rho\) of a conducting wire is the reciprocal of the conductivity and is measured in units of ohm-meters \((\Omega-m) .\) The resistivity of a given metal depends on the temperature according to the equation $$\rho(t)=\rho_{20} e^{\alpha(t-20)}$$ where \(t\) is the temperature in \(^{\circ} \mathrm{C}\) . where \(t\) is the temperature in \(^{\circ} \mathrm{C} .\) There are tables that list the values of \(\alpha\) (called the temperature coefficient) and \(\rho_{20}\) (the resistivity at \(20^{\circ} \mathrm{C} )\) for various metals. Except at very low temperatures, the resistivity varies almost linearly with temperature and so it is common to approximate the expression for \(\rho(t)\) by its first- or second-degree Taylor polynomial at \(t=20\) . (a) Find expressions for these linear and quadratic approximations. (b) For copper, the tables give \(\alpha=0.0039 /^{\circ} \mathrm{C}\) and \(\rho_{20}=1.7 \times 10^{-8} \Omega-\mathrm{m} .\) Graph the resistivity of copper and the linear and quadratic approximations for \(-250^{\circ} \mathrm{C} \leqslant t \leqslant 1000^{\circ} \mathrm{C}\) (c) For what values of \(t\) does the linear approximation agree with the exponential expression to within one percent?

(a) Show that the function $$f(x)=\sum_{n=0}^{\infty} \frac{x^{n}}{n !}$$ is a solution of the differential equation $$f^{\prime}(x)=f(x)$$ (b) Show that \(f(x)=e^{x}\)

Show that the sequence defined by $$ a_{1}=1 \quad a_{n+1}=3-\frac{1}{a_{n}} $$ is increasing and \(a_{n}<3\) for all \(n\) . Deduce that \(\left\\{a_{n}\right\\}\) is convergent and find its limit.

Find the sum of the series. $$\sum_{n=0}^{\infty} \frac{3^{n}}{5^{n} n !}$$
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