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Ch. 10 - Infinite Sequences and Series
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 10 - Infinite Sequences and SeriesProblem 10.6.10
Chapter 10, Problem 10.6.10

Determining Convergence or Divergence
In Exercises 1–14, determine whether the alternating series converges or diverges. Some of the series do not satisfy the conditions of the Alternating Series Test.
∑ (from n = 2 to ∞) [(-1)ⁿ⁺¹ (1 / ln n)]

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Identify the general term of the series: \(a_n = \frac{1}{\ln n}\) and the alternating factor is \((-1)^{n+1}\), so the series is \(\sum_{n=2}^\infty (-1)^{n+1} a_n\).
Recall the Alternating Series Test (Leibniz Test), which states that an alternating series \(\sum (-1)^n a_n\) converges if two conditions are met: (1) the sequence \(a_n\) is positive, decreasing, and (2) \(\lim_{n \to \infty} a_n = 0\).
Check if \(a_n = \frac{1}{\ln n}\) is positive and decreasing for \(n \geq 2\). Since \(\ln n\) is positive and increasing for \(n > 1\), \(a_n\) is positive and decreasing for \(n \geq 2\).
Evaluate the limit \(\lim_{n \to \infty} \frac{1}{\ln n}\). Since \(\ln n\) grows without bound as \(n\) approaches infinity, this limit is 0.
Since both conditions of the Alternating Series Test are satisfied, conclude that the series \(\sum_{n=2}^\infty (-1)^{n+1} \frac{1}{\ln n}\) converges.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Alternating Series Test

The Alternating Series Test determines convergence of series whose terms alternate in sign. It requires that the absolute value of terms decreases monotonically to zero. If these conditions hold, the series converges; otherwise, the test is inconclusive.
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Alternating Series Test

Behavior of the General Term

Analyzing the limit of the general term as n approaches infinity is crucial. For convergence, the terms must approach zero. In this series, the term 1/ln(n) decreases slowly, so checking its limit helps assess if the series can converge.
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Divergence Test (nth Term Test)

Logarithmic Functions and Their Growth

Understanding the growth rate of the natural logarithm function ln(n) is important. Since ln(n) grows without bound but very slowly, 1/ln(n) approaches zero but not rapidly. This slow decay affects the convergence behavior of the series.
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Graphs of Logarithmic Functions
Related Practice
Textbook Question

Telescoping Series

In Exercises 39–44, find a formula for the nth partial sum of the series and use it to determine if the series converges or diverges. If a series converges, find its sum.

∑ (from n = 1 to ∞) [ (1/n) − (1/(n + 1)) ]

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Textbook Question

Intervals of Convergence

In Exercises 1–36, (a) find the series’ radius and interval of convergence. For what values of x does the series converge (b) absolutely, (c) conditionally?

∑ (from n = 0 to ∞) (2x)ⁿ

Textbook Question

Are there any values of x for which ∑ (from n=1 to ∞) (1 / nˣ) converges? Give reasons for your answer.

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Textbook Question

Using the Ratio Test

In Exercises 1–8, use the Ratio Test to determine whether each series converges absolutely or diverges.

∑(from n=1 to ∞) [(-1)ⁿ (n + 2) / 3ⁿ]

Textbook Question

Using the Root Test

In Exercises 9–16, use the Root Test to determine if each series converges absolutely or diverges.

∑(from n=1 to ∞) [(-1)ⁿ (1 − 1/n)ⁿ^²]

(Hint: lim (n→∞) (1 + x/n)ⁿ = eˣ)

Textbook Question

Recursively Defined Sequences

In Exercises 101–108, assume that each sequence converges and find its limit.

a₁ = 2,aₙ₊₁ = 72 / (1 + aₙ)

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