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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

Here is a fascinating (unsolved) problem known as the hailstone problem (or the Ulam Conjecture or the Collatz Conjecture). It involves sequences in two different ways. First, choose a positive integer \(N\) and call it \(a_{0} .\) This is the seed of a sequence. The rest of the sequence is generated as follows: For \(n=0,1,2, \ldots\) $$a_{n+1}=\left\\{\begin{array}{ll} a_{n} / 2 & \text { if } a_{n} \text { is even } \\ 3 a_{n}+1 & \text { if } a_{n} \text { is odd. } \end{array}\right.$$ However, if \(a_{n}=1\) for any \(n,\) then the sequence terminates. a. Compute the sequence that results from the seeds \(N=2,3\), \(4, \ldots, 10 .\) You should verify that in all these cases, the sequence eventually terminates. The hailstone conjecture (still unproved) states that for all positive integers \(N,\) the sequence terminates after a finite number of terms. b. Now define the hailstone sequence \(\left\\{H_{k}\right\\},\) which is the number of terms needed for the sequence \(\left\\{a_{n}\right\\}\) to terminate starting with a seed of \(k\). Verify that \(H_{2}=1, H_{3}=7,\) and \(H_{4}=2\). c. Plot as many terms of the hailstone sequence as is feasible. How did the sequence get its name? Does the conjecture appear to be true?

Find the limit of the sequence $$\left\\{a_{n}\right\\}_{n=2}^{\infty}=\left\\{\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right) \cdots\left(1-\frac{1}{n}\right)\right\\}.$$

A well-known method for approximating \(\sqrt{c}\) for positive real numbers \(c\) consists of the following recurrence relation (based on Newton's method). Let \(a_{0}=c\) and $$a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{c}{a_{n}}\right), \quad \text { for } n=0,1,2,3, \dots$$ a. Use this recurrence relation to approximate \(\sqrt{10} .\) How many terms of the sequence are needed to approximate \(\sqrt{10}\) with an error less than \(0.01 ?\) How many terms of the sequence are needed to approximate \(\sqrt{10}\) with an error less than \(0.0001 ?\) (To compute the error, assume a calculator gives the exact value.) b. Use this recurrence relation to approximate \(\sqrt{c},\) for \(c=2\) \(3, \ldots, 10 .\) Make a table showing how many terms of the sequence are needed to approximate \(\sqrt{c}\) with an error less than \(0.01.\)

Pick two positive numbers \(a_{0}\) and \(b_{0}\) with \(a_{0}>b_{0}\) and write out the first few terms of the two sequences \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}:\) $$a_{n+1}=\frac{a_{n}+b_{n}}{2}, \quad b_{n+1}=\sqrt{a_{n} b_{n}}, \quad \text { for } n=0,1,2 \dots$$ (Recall that the arithmetic mean \(A=(p+q) / 2\) and the geometric mean \(G=\sqrt{p q}\) of two positive numbers \(p\) and \(q\) satisfy \(A \geq G\). a. Show that \(a_{n}>b_{n}\) for all \(n\). b. Show that \(\left\\{a_{n}\right\\}\) is a decreasing sequence and \(\left\\{b_{n}\right\\}\) is an increasing sequence. c. Conclude that \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}\) converge. d. Show that \(a_{n+1}-b_{n+1}<\left(a_{n}-b_{n}\right) / 2\) and conclude that \(\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n} .\) The common value of these limits is called the arithmetic-geometric mean of \(a_{0}\) and \(b_{0},\) denoted \(\mathrm{AGM}\left(a_{0}, b_{0}\right)\). e. Estimate AGM(12,20). Estimate Gauss' constant \(1 / \mathrm{AGM}(1, \sqrt{2})\).

Give an argument, similar to that given in the text for the harmonic series, to show that \(\sum_{k=1}^{\infty} \frac{1}{\sqrt{k}}\) diverges.

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