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Chapter 9: Problem 21

Find the limit of the following sequences or determine that the limit does not exist. $$\\{\sqrt{\left(1+\frac{1}{2 n}\right)^{n}}\\}$$

Short Answer

Expert verified
Answer: The limit of the sequence is 1.

Step by step solution

01

1. Write the sequence with limit notation

To begin, we'll write the given sequence as a limit function, as we're interested in finding the limit as n approaches infinity: $$\lim\limits_{n \to \infty} \sqrt{\left(1+\frac{1}{2 n}\right)^{n}}$$
02

2. Manipulate the expression inside the square root

We first focus on the expression inside the square root and look for a simplification: $$\lim\limits_{n \to \infty} \sqrt{\left(\frac{2n+1}{2n}\right)^{n}}$$
03

3. Apply the power property of the limit

We apply the power property of limits which states that if the limit of a function raised to a power exists, it's equal to the limit of the function, raised to that power: $$\left(\lim\limits_{n \to \infty}\frac{2n+1}{2n}\right)^{\lim\limits_{n \to \infty} n}$$
04

4. Calculate the limit of the ratio

As n approaches infinity, we only need to focus on the highest power of n in the numerator and the denominator to evaluate the limit of the ratio: $$\lim\limits_{n \to \infty}\frac{2n+1}{2n} = \frac{2}{2} = 1$$
05

5. Result of the limit

Now that we have evaluated the limit of the ratio, our expression becomes: $$1^{\lim\limits_{n \to \infty} n}$$
06

6. Analyze the exponent

We observe that the exponent approaches infinity, and the base of the power is equal to 1. By the properties of exponents, we know that for any non-zero number a, a raised to the power of infinity is infinity (unless a = 1). Since the base of our power function is 1, the exponent does not impact the result, any power of 1 remains 1: $$1^{\infty}=1$$
07

7. State the conclusion

To conclude, we find that the limit of the given sequence exists and has a value of 1: $$\lim\limits_{n \to \infty} \sqrt{\left(1+\frac{1}{2 n}\right)^{n}} = 1$$

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Most popular questions from this chapter

Suppose an alternating series \(\sum(-1)^{k} a_{k}\) converges to \(S\) and the sum of the first \(n\) terms of the series is \(S_{n}\) Suppose also that the difference between the magnitudes of consecutive terms decreases with \(k\). It can be shown that for \(n \geq 1,\) $$\left|S-\left[S_{n}+\frac{(-1)^{n+1} a_{n+1}}{2}\right]\right| \leq \frac{1}{2}\left|a_{n+1}-a_{n+2}\right|$$ a. Interpret this inequality and explain why it gives a better approximation to \(S\) than simply using \(S_{n}\) to approximate \(S\). b. For the following series, determine how many terms of the series are needed to approximate its exact value with an error less than \(10^{-6}\) using both \(S_{n}\) and the method explained in part (a). (i) \(\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k}\) (ii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{k \ln k}\) (iii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\sqrt{k}}\)

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 \(0.01.\)

Determine whether the following series converge or diverge. $$\sum_{k=0}^{\infty} \frac{k}{\sqrt{k^{2}+1}}$$

Evaluate the limit of the following sequences. $$a_{n}=\tan ^{-1}\left(\frac{10 n}{10 n+4}\right)$$

Evaluate the limit of the following sequences. $$a_{n}=\frac{7^{n}}{n^{7} 5^{n}}$$
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