Absolute and Conditional Convergence Calculator

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Example

Created on 2024-06-20Asked by Elijah Williams (Solvelet student)

Determine whether the series

\sum_{n=1}^{\infty} \frac{(-1)^n}{n \ln(n+1)}

is absolutely convergent, conditionally convergent, or divergent.

Solution

Step 1: Identify the series and the general term. The series is

\sum_{n=1}^{\infty} \frac{(-1)^n}{n \ln(n+1)}

with general term

a_n = \frac{(-1)^n}{n \ln(n+1)}.

Step 2: Check for Absolute Convergence. To check for absolute convergence, consider the series of absolute values:

\sum_{n=1}^{\infty} \left| \frac{(-1)^n}{n \ln(n+1)} \right| = \sum_{n=1}^{\infty} \frac{1}{n \ln(n+1)}.

Step 3: Apply the Integral Test for Absolute Convergence. Use the integral test by considering the function

f(x) = \frac{1}{x \ln(x+1)}

. Evaluate

\int_{1}^{\infty} \frac{1}{x \ln(x+1)} \, dx.

Step 4: Simplify the Integral. Let

u = \ln(x+1)

, then

du = \frac{1}{x+1} dx

. The limits change accordingly: when

x = 1

u = \ln(2)

; when

x \to \infty

u \to \infty

. The integral becomes

\int_{\ln(2)}^{\infty} \frac{1}{u} \, du.

Step 5: Evaluate the Integral. The integral

\int_{\ln(2)}^{\infty} \frac{1}{u} \, du

\left. \ln|u| \right|_{\ln(2)}^{\infty}.

This evaluates to

\infty

, indicating divergence. Thus,

\sum_{n=1}^{\infty} \frac{1}{n \ln(n+1)}

diverges. Step 6: Conclusion on Absolute Convergence. Since the series of absolute values diverges, the series

\sum_{n=1}^{\infty} \left| \frac{(-1)^n}{n \ln(n+1)} \right|

does not converge absolutely. Step 7: Check for Conditional Convergence. Since the series

\sum_{n=1}^{\infty} \frac{(-1)^n}{n \ln(n+1)}

is an alternating series, apply the Alternating Series Test (Leibniz's test). Step 8: Verify the Conditions for Alternating Series Test. The Alternating Series Test requires: (a) The terms

b_n = \frac{1}{n \ln(n+1)}

are positive. (b)

b_n

is decreasing. (c)

\lim_{n \to \infty} b_n = 0

. Step 9: Check if

b_n

is Positive and Decreasing. For

n \geq 1

b_n = \frac{1}{n \ln(n+1)}

is positive. To check if

b_n

is decreasing, consider

b_{n+1} = \frac{1}{(n+1) \ln(n+2)}

. Since

(n+1)\ln(n+2) > n \ln(n+1)

, we have

b_{n+1} < b_n

. Thus,

b_n

is decreasing. Step 10: Check the Limit. Calculate

\lim_{n \to \infty} b_n = \lim_{n \to \infty} \frac{1}{n \ln(n+1)} = 0.

Step 11: Conclusion on Conditional Convergence. Since

b_n

is positive, decreasing, and approaches zero as

n \to \infty

, the series

\sum_{n=1}^{\infty} \frac{(-1)^n}{n \ln(n+1)}

satisfies the conditions of the Alternating Series Test and is therefore conditionally convergent. Solved on Solvelet with Basic AI Model

Some of the related questions asked by Alexander Lewis on Solvelet

1. Determine whether the series

\sum_{n=1}^{\infty} (-1)^{n+1} / n

converges conditionally or absolutely 2. Find the interval of convergence for the power series

\sum_{n=0}^{\infty} \frac{(x-3)^n}{2^n}

DefinitionKnow & Understand absolute and conditional convergence: A series ∑ 𝑎 𝑛 ∑a n converges absolutely if ∑|𝑎 𝑛 | ∑|a n | converges. No matter the order, the series converges. A series ∑ 𝑎 𝑛 ∑a n converges conditionally if ∑ 𝑎 𝑛 ∑a n converges, but ∑|𝑎 𝑛 | ∑|a n | diverges. The series converges but changing the order changes the sum. Learn & solve absolute and conditional convergence with SolveletAI advanced step-by-step solutions. Generated instantly, learn and get explanations of the problem with absolute and conditional convergence calculator at SolveletAI

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