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Chapter 9: Problem 30

Use the Comparison Test or Limit Comparison Test to determine whether the following series converge. $$\sum_{k=1}^{\infty} \frac{0.0001}{k+4}$$

Short Answer

Expert verified
Answer: The series \(\sum_{k=1}^{\infty} \frac{0.0001}{k+4}\) diverges.

Step by step solution

01

Write down the given series and a comparison series

We are given the series: $$\sum_{k=1}^{\infty} \frac{0.0001}{k+4}$$ We will compare it to the harmonic series: $$\sum_{k=1}^{\infty} \frac{1}{k}$$
02

Find the limit of the ratio between the two series

We want to find the limit of the ratio between the given series and the harmonic series as k goes to infinity: $$\lim_{k\to\infty} \frac{\frac{0.0001}{k+4}}{\frac{1}{k}}$$
03

Simplify the limit expression

Now, let's simplify the limit expression: $$\lim_{k\to\infty} \frac{\frac{0.0001}{k+4}}{\frac{1}{k}} = \lim_{k\to\infty} \frac{0.0001k}{k+4}$$
04

Evaluate the limit

We can now evaluate the limit as k goes to infinity: $$\lim_{k\to\infty} \frac{0.0001k}{k+4} = \lim_{k\to\infty} \frac{0.0001}{1+\frac{4}{k}}$$ Notice that the term \(\frac{4}{k}\) goes to 0 as k goes to infinity, thus the limit becomes: $$\lim_{k\to\infty} \frac{0.0001}{1+\frac{4}{k}} = \frac{0.0001}{1+0} = 0.0001$$
05

Apply the Limit Comparison Test

Since we have found that the limit is equal to a positive constant (0.0001), the Limit Comparison Test tells us that the series we started with, \(\sum_{k=1}^{\infty} \frac{0.0001}{k+4}\), has the same convergence behavior as the harmonic series \(\sum_{k=1}^{\infty} \frac{1}{k}\).
06

Determine the convergence of the given series

We know that the harmonic series is a divergent series. Since both the given series and the harmonic series have the same convergence behavior, the given series: $$\sum_{k=1}^{\infty} \frac{0.0001}{k+4}$$ also diverges.

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Most popular questions from this chapter

Convergence parameter Find the values of the parameter \(p>0\) for which the following series converge. $$\sum_{k=2}^{\infty} \frac{1}{(\ln k)^{p}}$$

A glimpse ahead to power series Use the Ratio Test to determine the values of \(x \geq 0\) for which each series converges. $$\sum_{k=1}^{\infty} \frac{x^{k}}{k^{2}}$$

Suppose a function \(f\) is defined by the geometric series \(f(x)=\sum_{k=0}^{\infty} x^{2 k}\) a. Evaluate \(f(0), f(0.2), f(0.5), f(1),\) and \(f(1.5),\) if possible. b. What is the domain of \(f ?\)

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

a. Sketch the function \(f(x)=1 / x\) on the interval \([1, n+1]\) where \(n\) is a positive integer. Use this graph to verify that $$ \ln (n+1)<1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}<1+\ln n $$ b. Let \(S_{n}\) be the sum of the first \(n\) terms of the harmonic series, so part (a) says \(\ln (n+1)0,\) for \(n=1,2,3, \ldots\) c. Using a figure similar to that used in part (a), show that $$ \frac{1}{n+1}>\ln (n+2)-\ln (n+1) $$ d. Use parts (a) and (c) to show that \(\left\\{E_{n}\right\\}\) is an increasing sequence \(\left(E_{n+1}>E_{n}\right)\) e. Use part (a) to show that \(\left\\{E_{n}\right\\}\) is bounded above by 1 f. Conclude from parts (d) and (e) that \(\left\\{E_{n}\right\\}\) has a limit less than or equal to \(1 .\) This limit is known as Euler's constant and is denoted \(\gamma\) (the Greek lowercase gamma). g. By computing terms of \(\left\\{E_{n}\right\\},\) estimate the value of \(\gamma\) and compare it to the value \(\gamma \approx 0.5772 .\) (It has been conjectured, but not proved, that \(\gamma\) is irrational.) h. The preceding arguments show that the sum of the first \(n\) terms of the harmonic series satisfy \(S_{n} \approx 0.5772+\ln (n+1)\) How many terms must be summed for the sum to exceed \(10 ?\)
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