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Chapter 9: Problem 24

Find the limit of the following sequences or determine that the limit does not exist. $$\left\\{\frac{\ln (1 / n)}{n}\right\\}$$

Short Answer

Expert verified
Answer: The limit of the sequence as \(n\) approaches infinity is \(0\).

Step by step solution

01

Rewrite the sequence with limit notation

We want to find the limit as \(n\) approaches infinity. So, we write: $$\lim_{n\to\infty} \frac{\ln\left(\frac{1}{n}\right)}{n}$$
02

Express the logarithm as a natural log

Recall that \(\ln\left(\frac{1}{n}\right)\) is the natural logarithm of \(\frac{1}{n}\), which can also be written as \(-\ln(n)\). So, the limit becomes: $$\lim_{n\to\infty} \frac{-\ln(n)}{n}$$
03

Check L'Hopital's rule conditions

We can see that both the numerator and denominator tend to infinity as \(n\) approaches infinity. L'Hopital's rule can be applied when the limit of both the numerator and denominator is either zero or infinity. Since we have infinity in this case, we can proceed.
04

Apply L'Hopital's rule

Differentiate the numerator and denominator with respect to \(n\): $$\frac{d}{dn}(-\ln(n)) = -\frac{1}{n}$$ $$\frac{d}{dn}(n) = 1$$ Now, rewrite the limit with the new numerator and denominator: $$\lim_{n\to\infty} \frac{-\frac{1}{n}}{1}$$
05

Evaluate the limit

To evaluate the limit, we divide the numerator by the denominator as \(n\) approaches infinity: $$\lim_{n\to\infty} \left(-\frac{1}{n}\right) = 0$$ Thus, the limit of the sequence \(\left\\{\frac{\ln (1 / n)}{n}\right\\}\) as \(n\) approaches infinity is \(0\).

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Most popular questions from this chapter

Does a geometric series always have a finite value?

Consider the formulas for the following sequences. Using a calculator, make a table with at least 10 terms and determine a plausible value for the limit of the sequence or state that it does not exist. $$a_{n}=\frac{(n-1)^{2}}{\left(n^{2}-1\right)} ; n=2,3,4, \dots$$

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

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 \(1 / \mathrm{AGM}(1, \sqrt{2})\).

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