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Chapter 7: Problem 59

Use the Limit Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{n^{k-1}}{n^{k}+1}, \quad k>2 $$

Short Answer

Expert verified
The series \( \frac{n^{k-1}}{n^{k}+1} \) converges.

Step by step solution

01

Choose a Comparable Series

Choose a series that is easy to compare with the given series. In this case, choose \( b_n = \frac{1}{n^2} \).
02

Calculate the Limit

Calculate the limit as \( n \) approaches infinity of \( \frac{a_n}{b_n} \). We have,\[ \lim_{n\to\infty} \frac{n^{k-1}}{n^{k}+1} \cdot n^2 = \lim_{n\to\infty} \frac{n^{k-1+2}}{n^{k}+1} = \lim_{n\to\infty} \frac{n^{k+1}}{n^{k}+1}= 0 \] where \( a_n = \frac{n^{k-1}}{n^{k}+1} \) and \( b_n = \frac{1}{n^2} \).
03

Apply the Limit Comparison Test

Applying the limit comparison test, since the limit is 0 and \( k > 2 \), scenarios wherein the limit as \( n \) approaches infinity of \( \frac{a_n}{b_n} \) is some finite number (>0) or zero implies that both \( a_n \) and \( b_n \) will either both converge or both diverge. Since \( b_n = \frac{1}{n^2} \) is a p-series with \( p = 2 > 1 \), then \( b_n \) is convergent. Therefore, \( a_n = \frac{n^{k-1}}{n^{k}+1} \) is also convergent.

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Most popular questions from this chapter

Prove that \(\frac{1}{r}+\frac{1}{r^{2}}+\frac{1}{r^{3}}+\cdots=\frac{1}{r-1}\) for \(|r|>1\).

Find the sum of the convergent series. $$ \sum_{n=1}^{\infty}\left[(0.7)^{n}+(0.9)^{n}\right] $$

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 \(x\). $$ \sum_{n=0}^{\infty} 4\left(\frac{x-3}{4}\right)^{n} $$

Compute the first six terms of the sequence \(\left\\{a_{n}\right\\}=\left\\{\left(1+\frac{1}{n}\right)^{n}\right\\}\) If the sequence converges, find its limit.

State the definitions of convergent and divergent series.
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