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Limit Comparison Test Calculator

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Example
Created on 2024-06-20Asked by Aria Williams (Solvelet student)
Use the limit comparison test to determine whether the series ∑n=1∞3n2+2n3+1 \sum_{n=1}^{\infty} \frac{3n^2 + 2}{n^3 + 1} converges or diverges by comparing it with ∑n=1∞1n \sum_{n=1}^{\infty} \frac{1}{n} .

Solution

We use the limit comparison test to determine whether the series ∑n=1∞3n2+2n3+1 \sum_{n=1}^{\infty} \frac{3n^2 + 2}{n^3 + 1} converges or diverges by comparing it with the harmonic series ∑n=1∞1n \sum_{n=1}^{\infty} \frac{1}{n} . Consider an=3n2+2n3+1 a_n = \frac{3n^2 + 2}{n^3 + 1} and bn=1n b_n = \frac{1}{n} . Calculate the limit: lim⁡n→∞anbn=lim⁡n→∞3n2+2n3+11n=lim⁡n→∞3n3+2nn3+1=lim⁡n→∞3+2n21+1n3=3 \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{3n^2 + 2}{n^3 + 1}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{3n^3 + 2n}{n^3 + 1} = \lim_{n \to \infty} \frac{3 + \frac{2}{n^2}}{1 + \frac{1}{n^3}} = 3 Since the limit is a positive finite number (3), and since the harmonic series ∑n=1∞1n \sum_{n=1}^{\infty} \frac{1}{n} diverges, the given series ∑n=1∞3n2+2n3+1 \sum_{n=1}^{\infty} \frac{3n^2 + 2}{n^3 + 1} also diverges by the limit comparison test. Solved on Solvelet with Basic AI Model
Some of the related questions asked by Madison Anderson on Solvelet
1. Use the limit comparison test to determine the convergence or divergence of the series ∑n2+3n4−1\sum \frac{n^2 + 3}{n^4 - 1},2. Find the limit lim⁡x→0ex−1x\lim_{x \to 0} \frac{e^x - 1}{x}.,
DefinitionA Limit Comparison Test Calculator is used to find out whether a given infinite series converges or diverges to infinity by comparing the infinite series with a series that is known to converge or diverges to infinity. The limit of the ratio of the two series is calculated. Example: To test the convergence of ( \sum\frac{1}{n^2+1} ), you may compare it with ( \sum\frac{1}{n^2} ). The orders that the calculator calculates in the numerator would be the limit \[ \lim_{{n \to \infty}} \frac{\frac{1}{n^2+1}}{\frac{1}{n^2}} = \lim_{{n \to \infty}} \frac{n^2}{n^2+1} = 1\] If the limit is a finite non-zero number - both series converge or diverge; Since ( \sum \frac{1}{n^2} ) is convergent, ( \sum \frac{1}{n^2+1} ) is also convergent.
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