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Chapter 9: Problem 71

Use the formal definition of the limit of a sequence to prove the following limits. $$\lim _{n \rightarrow \infty} \frac{3 n^{2}}{4 n^{2}+1}=\frac{3}{4}$$

Short Answer

Expert verified
Question: Prove that the limit of the sequence \(a_n = \frac{3n^2}{4n^2+1}\) as n approaches infinity is \(\frac{3}{4}\). Answer: We have proven that the limit of the sequence \(a_n = \frac{3n^2}{4n^2+1}\) as n approaches infinity is \(\frac{3}{4}\), using the formal definition of a limit, by finding a suitable value for N depending on \(\epsilon > 0\) such that for all n > N, the absolute difference between the terms of the sequence and the limit is less than \(\epsilon\).

Step by step solution

01

Express the difference between the terms and the limit

Let's first express the difference between the terms of the sequence and the limit, which is given by: $$\left| \frac{3 n^{2}}{4 n^{2}+1} - \frac{3}{4} \right|$$
02

Find a common denominator

In order to work with this expression, we'll need to find a common denominator for the two fractions. To do this, we multiply the second fraction by \(\frac{4n^2+1}{4n^2+1}\), giving us: $$\left| \frac{3 n^{2}}{4 n^{2}+1} - \frac{3(4n^2+1)}{4(4n^2+1)} \right|$$
03

Combine the fractions

Now, we can combine the two fractions by subtracting their numerators: $$\left| \frac{3n^2 - 3(4n^2+1)}{4(4n^2+1)} \right|$$
04

Simplify the numerator

Next, let's simplify the expression in the numerator: $$\left| \frac{3n^2 - 12n^2 - 3}{4(4n^2+1)} \right| = \left| \frac{-9n^2 - 3}{4(4n^2+1)} \right|$$
05

Determine a bound for the expression

To get a bound for this expression, we can divide the absolute value of the numerator by the absolute value of the denominator: $$\frac{9n^2 + 3}{4(4n^2+1)}$$ For n > 0, we have that: $$0 < \frac{9n^2 + 3}{4(4n^2+1)} < \frac{9n^2+9n^2}{4(4n^2)} = \frac{18n^2}{16n^2} = \frac{9}{8}$$
06

Find a suitable value for N

Now, we want to find N such that for all n > N, we have: $$\frac{9n^2 + 3}{4(4n^2+1)} < \epsilon$$ We already have: $$\frac{9n^2 + 3}{4(4n^2+1)} < \frac{9}{8}$$ So, if we can find an N satisfying: $$\frac{9}{8} < \epsilon$$ Then, for all n > N, we will have: $$\frac{9n^2 + 3}{4(4n^2+1)} < \epsilon$$ Let's take N as the smallest integer greater than or equal to \(\frac{9}{8\epsilon}\). Since this N depends on \(\epsilon\), for every \(\epsilon > 0\), we can find an appropriate N. In conclusion, we have proven that: $$\lim _{n \rightarrow \infty} \frac{3 n^{2}}{4 n^{2}+1}=\frac{3}{4}$$ by the formal definition of a limit.

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Most popular questions from this chapter

Consider the series \(\sum_{k=3}^{\infty} \frac{1}{k \ln k(\ln \ln k)^{p}},\) where \(p\) is a real number. a. For what values of \(p\) does this series converge? b. Which of the following series converges faster? Explain. $$ \sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{2}} \text { or } \sum_{k=3}^{\infty} \frac{1}{k \ln k(\ln \ln k)^{2}} ? $$

Determine whether the following statements are true and give an explanation or counterexample. a. A series that converges must converge absolutely. b. A series that converges absolutely must converge. c. A series that converges conditionally must converge. d. If \(\sum a_{k}\) diverges, then \(\Sigma\left|a_{k}\right|\) diverges. e. If \(\sum a_{k}^{2}\) converges, then \(\sum a_{k}\) converges. f. If \(a_{k}>0\) and \(\sum a_{k}\) converges, then \(\Sigma a_{k}^{2}\) converges. g. If \(\Sigma a_{k}\) converges conditionally, then \(\Sigma\left|a_{k}\right|\) diverges.

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

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

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. $$a_{n+1}=\frac{1}{2} a_{n}+2 ; a_{0}=5, n=0,1,2, \dots$$
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