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Chapter 7: Problem 54

Use the Limit Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=3}^{\infty} \frac{3}{\sqrt{n^{2}-4}} $$

Short Answer

Expert verified
The given series diverges.

Step by step solution

01

Identify the comparable series

An obvious choice here as a comparison series would be \( \frac{1}{n} \) since it is simpler but still resembles the original series\( \frac{3}{\sqrt{n^{2}-4}} \). The series \( \frac{1}{n} \) is the harmonic series and it is known to diverge.
02

Apply the limit comparison test

The Limit Comparison Test says that if \( \lim_{{n \to \infty}} \frac{a_{n}}{b_{n}} = c > 0 \), then the series \(\sum_{n=1}^{\infty}a_{n}\) and \(\sum_{n=1}^{\infty}b_{n}\) either both converge or both diverge. Here \(a_{n} = \frac{3}{\sqrt{n^{2}-4}}\) and \(b_{n} = \frac{1}{n}\). So compute the limit, \( \lim_{{n \to \infty}} \frac{a_{n}}{b_{n}} = \lim_{{n \to \infty}} \frac{n*3}{\sqrt{n^{2}-4}}. \)
03

Simplify the limit

The fraction under the limit simplifies to \(3\frac{n}{\sqrt{n^{2}-4}}\), and taking limit as \(n \to \infty\), it approaches 3. Since the limit is a positive constant, according to the Limit Comparison Test, the given series has the same behavior as the series we compared it with.
04

Apply the result of the Limit Comparison Test

The comparison series \( \frac{1}{n} \) is a diverging series (Harmonic series) and since we applied the Limit Comparison Test, it means that our original series \( \frac{3}{\sqrt{n^{2}-4}} \) also diverges.

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Most popular questions from this chapter

Find all values of \(x\) for which the series converges. For these values of \(x,\) write the sum of the series as a function of \(x\). $$ \sum_{n=0}^{\infty}(-1)^{n} x^{n} $$

The Fibonacci sequence is defined recursively by \(a_{n+2}=a_{n}+a_{n+1},\) where \(a_{1}=1\) and \(a_{2}=1\) (a) Show that \(\frac{1}{a_{n+1} a_{n+3}}=\frac{1}{a_{n+1} a_{n+2}}-\frac{1}{a_{n+2} a_{n+3}}\). (b) Show that \(\sum_{n=0}^{\infty} \frac{1}{a_{n+1} a_{n+3}}=1\).

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

(a) write the repeating decimal as a geometric series and (b) write its sum as the ratio of two integers $$ 0.2 \overline{15} $$

State the \(n\) th-Term Test for Divergence.
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