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Test the following series for convergence. $$\sum_{n=2}^{\infty} \frac{(-1)^{n}}{\ln n}$$

Short Answer

Expert verified
The series converges by the Alternating Series Test.

Step by step solution

01

Identify the series

Identify the given series: the series is the alternating series $$\frac{(-1)^{n}}{\text{ln} n}$$
02

Apply the Alternating Series Test conditions

The Alternating Series Test (Leibniz test) states that a series of the form $$\text{$$\text{$$\text{$$\text{\text{'Leibniz test'}{-1}}^{n}b_{n}$$$$will converge if:$$1.$$ The terms $$\text{b}_{n}$$ decrease monotonically: \text{ \text{if \text{}{ln }): The natural logarithm function, uniformly }ormally increases. To prove the terms decrease, rewrite as: $$b_{n} = \frac{1}{\text{ln} {n}}$$
03

Check if $$1/\text{ln}$$ n decreases

Show $$1/\text{ln}$$ n is decreasing. Consider the derivative of $$ b_{n} = \frac{1}{\text{ln}{n}}: interpret as decreasing: \text{1:$$ - (\text{ln}(\text{}{n} is decreasing for} {n\text=>=2}.$$
04

Check the limit condition

Evaluate $$\text{}$$ Evaluate the limit condition:$$$\text{If both are}$$\text{}else converging:\text{=0}):$$$Step (The series fulfills): These conditions:: The limit conditions have been fulfilled):)$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Series Convergence
In mathematics, a series is the sum of the terms of a sequence of numbers. To determine if a series converges means figuring out if adding the sequence's terms approaches a specific value as more terms are included. Understanding series convergence is crucial because it tells us if an infinite sum can result in a finite value.
Leibniz Test
The Leibniz test, also known as the Alternating Series Test, is used to check if an alternating series converges. An alternating series alternates between positive and negative terms. To apply the Leibniz test, two main conditions need to be satisfied:
Natural Logarithm
The natural logarithm, denoted as \(\text{ln}(x)\), is the logarithm to the base \(e\) (where \(e \approx 2.71828\)). It helps us understand growth processes, such as compound interest. Understanding the properties of natural logarithms is helpful in analyzing functions like those seen in series convergence tests.

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Most popular questions from this chapter

Test the following series for convergence or divergence. Decide for yourself which test is easiest to use, but don't forget the preliminary test. Use the facts stated above when they apply. $$\sum_{n=1}^{\infty} \frac{n^{n}}{n !}$$

Test for convergence: $$\sum_{n=1}^{\infty} \frac{2^{n}}{n !}$$

Test for convergence: $$\sum_{n=2}^{\infty} \frac{1}{n \ln \left(n^{3}\right)}$$

(a) It is clear that you (or your computer) can't find the sum of an infinite series just by adding up the terms one by one. For example, to get \(\zeta(1.1)=\) \(\sum_{n=1}^{\infty} 1 / n^{1.1}(\text { see Problem } 15.22)\) with error \(<0.005\) takes about \(10^{33}\) terms. To see a simple alternative (for a series of positive decreasing terms) look at Figures 6.1 and 6.2. Show that when you have summed \(N\) terms, the sum \(R_{N}\) of the rest of the series is between \(I_{N}=\int_{N}^{\infty} a_{n} d n\) and \(I_{N+1}=\int_{N+1}^{\infty} a_{n} d n\) (b) Find the integrals in (a) for the \(\zeta(1.1)\) series and verify the claimed number of terms needed for error \(<0.005 .\) Hint: Find \(N\) such that \(I_{N}=0.005 .\) Also find upper and lower bounds for \(\zeta(1.1)\) by computing \(\sum_{n=1}^{N} 1 / n^{1.1}+\int_{N}^{\infty} n^{-1.1} d n\) and \(\sum_{n=1}^{N} 1 / n^{1.1}+\int_{N+1}^{\infty} n^{-1.1} d n\) where \(N\) is far less than \(10^{33} .\) Hint: You want the difference between the upper and lower limits to be about \(0.005 ;\) find \(\mathrm{N}\) so that term \(a_{N}=0.005\).

Find the interval of convergence of each of the following power series; be sure to investigate the endpoints of the interval in each case. $$\sum_{n=1}^{\infty} \frac{(x-2)^{n}}{3^{n}}$$

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