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Chapter 7: Problem 45

Use the Direct Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=2}^{\infty} \frac{\ln n}{n+1} $$

Short Answer

Expert verified
By using the Direct Comparison Test and comparing the given series to the harmonic series, we can conclude that the given series \( \sum_{n=2}^{\infty} \frac{\ln n}{n+1} \) is divergent.

Step by step solution

01

Understand the Direct Comparison Test

The Direct Comparison Test is a method to determine if an infinite series converges or diverges. The idea is that if series a_n is known to converge or diverge, then a comparison series b_n can be found such that \(0 \leq a_n \leq b_n\) or \(0 \leq b_n \leq a_n \) for all n. If a_n converges and \(a_n \leq b_n\) for all n, then b_n also converges. If a_n diverges and \(b_n \leq a_n\) for all n, then b_n also diverges.
02

Establish a Comparison Series

The harmonic series \( \sum_{n=2}^{\infty} \frac {1}{n} \) is a known divergent series which seems like it could be used for comparison with the original series \( \sum_{n=2}^{\infty} \frac{\ln n}{n+1} \). Let's evaluate and compare the terms of these series for large n.
03

Comparing terms of the series

Consider the inequality \(0 \leq \frac {1}{n} \leq \frac {\ln n}{n+1}\) for \(n \geq 2\). We can see that both inequalities are true: the term on the left is always positive and as n goes to infinity, the fractions are decreasing. Using the Direct Comparison Test, since the harmonic series \( \frac {1}{n} \) is diverging, the original series \( \frac {\ln n}{n+1} \) will also diverge because its terms are greater than the terms of a known diverging series.

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Most popular questions from this chapter

(a) You delete a finite number of terms from a divergent series. Will the new series still diverge? Explain your reasoning. (b) You add a finite number of terms to a convergent series. Will the new series still converge? Explain your reasoning.

Compound Interest Consider the sequence \(\left\\{A_{n}\right\\}\) whose \(n\) th term is given by \(A_{n}=P\left(1+\frac{r}{12}\right)^{n}\) where \(P\) is the principal, \(A_{n}\) is the account balance after \(n\) months, and \(r\) is the interest rate compounded annually. (a) Is \(\left\\{A_{n}\right\\}\) a convergent sequence? Explain. (b) Find the first 10 terms of the sequence if \(P=\$ 9000\) and \(r=0.055\)

(a) write the repeating decimal as a geometric series and (b) write its sum as the ratio of two integers $$ 0 . \overline{81} $$

Find the sum of the convergent series. $$ 4-2+1-\frac{1}{2}+\cdots $$

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