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Chapter 9: Problem 13

Find the limit of the following sequences or determine that the limit does not exist. $$\left\\{\frac{3^{n+1}+3}{3^{n}}\right\\}$$

Short Answer

Expert verified
Answer: The limit of the sequence as n approaches infinity for the given expression is 3.

Step by step solution

01

Factor out the common term

We will first factor out 3^n from the expression: $$\frac{3^{n+1}+3}{3^{n}} = \frac{3^n\cdot3+3}{3^n}$$
02

Simplify the expression

Now we can simplify the expression by dividing both terms in the numerator by \(3^n\): $$\frac{3^n\cdot3+3}{3^n} = \frac{3^n\cdot3}{3^n}+\frac{3}{3^n} = 3+\frac{3}{3^n}$$
03

Take the limit as n approaches infinity

As n approaches infinity, let's analyze each term of the expression: - The first term, 3, remains constant and does not change. - The second term, \(\frac{3}{3^n}\), approaches 0 as n approaches infinity, since the denominator grows without bound.
04

Determine the limit

Based on our analysis in Step 3, we can conclude that as n approaches infinity, the sequence converges to the value 3. Therefore, the limit of the sequence is 3. The final result is: $$\lim_{n\to\infty} \left\\{\frac{3^{n+1}+3}{3^{n}}\right\\} = 3$$

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Most popular questions from this chapter

The prime numbers are those positive integers that are divisible by only 1 and themselves (for example, 2,3,5 7, \(11,13, \ldots .\) ). A celebrated theorem states that the sequence of prime numbers \(\left\\{p_{k}\right\\}\) satisfies \(\lim _{k \rightarrow \infty} p_{k} /(k \ln k)=1 .\) Show that \(\sum_{k=2}^{\infty} \frac{1}{k \ln k}\) diverges, which implies that the series \(\sum_{k=1}^{\infty} \frac{1}{p_{k}}\) diverges.

Determine whether the following series converge absolutely or conditionally, or diverge. $$\sum_{k=1}^{\infty} \frac{(-1)^{k} \tan ^{-1} k}{k^{3}}$$

Showing that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6} \operatorname{In} 1734,\) Leonhard Euler informally proved that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6} .\) An elegant proof is outlined here that uses the inequality $$ \cot ^{2} x<\frac{1}{x^{2}}<1+\cot ^{2} x\left(\text { provided that } 0

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 absolutely or conditionally, or diverge. $$\sum_{k=1}^{\infty}\left(-\frac{1}{3}\right)^{k}$$
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