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Chapter 7: Problem 89

Compute the first six terms of the sequence \(\left\\{a_{n}\right\\}=\\{\sqrt[n]{n}\\} .\) If the sequence converges, find its limit.

Short Answer

Expert verified
The first six terms of the sequence are approximately 1, 1.414, 1.442, 1.414, 1.379, 1.348. The sequence does converge, and the limit is 1.

Step by step solution

01

Calculate the first six terms

This step involves calculating the first six terms of the sequence. So, substitute 1, 2, 3, 4, 5, and 6 into \(a_n\) = \(\sqrt[n]{n}\).The first six terms are then:1. \(\sqrt[1]{1} = 1\)2. \(\sqrt[2]{2} = \sqrt{2} \approx 1.414\)3. \(\sqrt[3]{3} \approx 1.442\)4. \(\sqrt[4]{4} = \sqrt{2} \approx 1.414\)5. \(\sqrt[5]{5} \approx 1.379\)6. \(\sqrt[6]{6} \approx 1.348\)
02

Determine if the sequence converges

From our calculations, we observe the values of the terms appear to decrease beyond the first term. They are indeed getting closer and closer to some finite number, and hence, one could assume the sequence converges.
03

Find the limit of the sequence

Informally, the limit of a sequence is the value that the terms of a sequence 'get closer and closer to' as the index increases without bound. We need to apply a limit operation to our sequence \(a_n = \sqrt[n]{n}\). The formal limit is denoted as: \[\lim_{n \to \infty} \sqrt[n]{n}\] Using the limit laws, this equation can be rewritten as: \[\lim_{n \to \infty} n^{1/n}\] And when we apply the rule that \[\lim_{n \to \infty} n^{1/n} = 1\] it gives us the limit of 1. Logarithms can be used to prove this limit. So, if the sequence \(\{a_{n}\}\) converges, it converges to 1.

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Most popular questions from this chapter

Show that the series \(\sum_{n=1}^{\infty} a_{n}\) can be written in the telescoping form \(\sum_{n=1}^{\infty}\left[\left(c-S_{n-1}\right)-\left(c-S_{n}\right)\right]\) where \(S_{0}=0\) and \(S_{n}\) is the \(n\) th partial sum.

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(n>1\), then \(n !=n(n-1) !\)

Find the sum of the convergent series. $$ \sum_{n=0}^{\infty}\left(-\frac{1}{2}\right)^{n} $$

Prove that if \(\left\\{s_{n}\right\\}\) converges to \(L\) and \(L>0,\) then there exists a number \(N\) such that \(s_{n}>0\) for \(n>N\).

Find the sum of the convergent series. $$ \sum_{n=0}^{\infty} 2\left(-\frac{2}{3}\right)^{n} $$
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