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Chapter 9: Problem 61

Choose your test Use the test of your choice to determine whether the following series converge. $$\sum_{k=1}^{\infty} \ln \left(\frac{k+2}{k+1}\right)$$

Short Answer

Expert verified
If it converges, what is its sum? Answer: The given series converges, with the sum being \(\ln(3)\).

Step by step solution

01

Simplify the expression

Let's simplify the terms involved in the series using logarithmic properties. According to the logarithmic properties, we have: $$\ln \left(\frac{k+2}{k+1}\right) = \ln(k+2) - \ln(k+1)$$ Now, the given series becomes: $$\sum_{k=1}^{\infty} (\ln(k+2) - \ln(k+1))$$
02

Rewrite the series as a telescoping series

Notice that the summation has a cancellation pattern where each term is cancelling out with its corresponding term: $$\left(\ln(3) - \ln(2)\right) + \left(\ln(4) - \ln(3)\right) + \left(\ln(5) - \ln(4)\right) + \cdots$$ This is a telescoping series, which can be written as: $$\left(\ln(3) - \ln(2)\right) + \left(\ln(4) - \ln(3)\right) + \left(\ln(5) - \ln(4)\right) + \cdots = (\ln(3) + \ln(4) + \cdots ) - \ln(2) - \ln(3) -\ln(4) - \cdots$$
03

Evaluate the series

Since the series is telescoping, the subsequent terms are cancelling out except for the first term. So, the sum of this series is: $$\sum_{k=1}^{\infty} \ln \left(\frac{k+2}{k+1}\right) = \ln(3)$$ Since \(\ln(3)\) is finite, we can conclude that the series converges. Therefore, the given series converges, with the sum being \(\ln(3)\).

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Most popular questions from this chapter

Determine whether the following statements are true and give an explanation or counterexample. a. A series that converges must converge absolutely. b. A series that converges absolutely must converge. c. A series that converges conditionally must converge. d. If \(\sum a_{k}\) diverges, then \(\Sigma\left|a_{k}\right|\) diverges. e. If \(\sum a_{k}^{2}\) converges, then \(\sum a_{k}\) converges. f. If \(a_{k}>0\) and \(\sum a_{k}\) converges, then \(\Sigma a_{k}^{2}\) converges. g. If \(\Sigma a_{k}\) converges conditionally, then \(\Sigma\left|a_{k}\right|\) diverges.

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

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

Use Exercise 89 to determine how many terms of each series are needed so that the partial sum is within \(10^{-6}\) of the value of the series (that is, to ensure \(R_{n}<10^{-6}\) ). a. \(\sum_{k=0}^{\infty} 0.6^{k}\) b. \(\sum_{k=0}^{\infty} 0.15^{k}\)

Infinite products Use the ideas of Exercise 88 to evaluate the following infinite products. $$\text { a. } \prod_{k=0}^{\infty} e^{1 / 2^{k}}=e \cdot e^{1 / 2} \cdot e^{1 / 4} \cdot e^{1 / 8} \ldots$$ $$\text { b. } \prod_{k=2}^{\infty}\left(1-\frac{1}{k}\right)=\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \dots$$
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