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Chapter 9: Problem 18

Find the limit of the following sequences or determine that the limit does not exist. $$\left\\{n^{2 / n}\right\\}$$

Short Answer

Expert verified
" The short answer would be: "The limit of the sequence is 1."

Step by step solution

01

Sequence Definition

We are given the sequence: $$a_n = n^{2/n}$$ Our goal is to find the limit of this sequence as \(n \to \infty\), or determine that the limit does not exist.
02

Rewrite the Sequence in Exponential Form

In order to evaluate the limit, we can rewrite the sequence using an equivalent exponential form: $$a_n = e^{\ln(n^{2/n})}$$ This is because \(n^{2/n} = e^{\ln(n^{2/n})}\) due to the property: \(x = e^{\ln(x)}\)
03

Apply Logarithm Properties

Now, we can apply logarithm properties to simplify the expression within the exponent: $$a_n = e^{\frac{2}{n}\ln(n)}$$ The property used is \(\ln(a^b) = b \cdot \ln(a)\)
04

Find the Limit

We want to find the limit of the sequence as \(n \to \infty\): $$\lim_{n\to\infty} a_n = \lim_{n\to\infty} e^{\frac{2}{n}\ln(n)}$$ To solve this limit, we can use L'H么pital's Rule. First, let's rewrite the limit as the indeterminate form \(\frac{0}{0}\) by considering the limit of \(\frac{\ln(n)}{n}\): $$\lim_{n\to\infty} \frac{2\ln(n)}{n}$$
05

Apply L'H么pital's Rule

Now, we can apply L'H么pital's Rule by differentiating the numerator and denominator with respect to n: $$\lim_{n\to\infty} \frac{\frac{2}{n}}{1} = \lim_{n\to\infty} \frac{2}{n}$$ As n approaches infinity, \(\frac{2}{n}\) approaches 0: $$\lim_{n\to\infty} \frac{2}{n} = 0$$
06

Find the Original Limit

With the result from Step 5, we can now find the original limit: $$\lim_{n\to\infty} a_n = \lim_{n\to\infty} e^{\frac{2}{n}\ln(n)} = e^{\lim_{n\to\infty} \frac{2\ln(n)}{n}} = e^0$$ Since \(e^0 = 1\), the limit of the sequence is 1: $$\lim_{n\to\infty} a_n = 1$$ Thus, the limit of the sequence $$\left\\{n^{2 / n}\right\\}$$ as n approaches infinity is 1.

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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. $$\sum_{k=2}^{\infty} \frac{1}{k \ln k(\ln \ln k)^{p}}$$

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

Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the value of the limit or determine that the limit does not exist. $$a_{n+1}=\frac{1}{2} a_{n}+2 ; a_{0}=5, n=0,1,2, \dots$$

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 \(10 ?\)

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?
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