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Limit Comparison Test for Series Convergence

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Use the limit comparison test to determine if the series n2n5+8\frac{n^2}{n^5 + 8} converges or diverges.

The limit comparison test is a core component of series analysis, especially when dealing with complex rational expressions. This test is grounded on the concept of comparing a given series to another series that is known to converge or diverge, thereby gauging the behavior of the original series indirectly. The key idea behind this technique is to find a comparable series whose convergence properties are known and is of a form that behaves similarly to the series in question for large values of n.

When analyzing the series n2n5+8\frac{n^2}{n^5 + 8}, observe that the dominant term in the denominator, n5n^5, suggests that a simpler comparison series such as 1n3\frac{1}{n^3} could be relevant, as it is derived by simplifying the terms based on their highest power. Understanding the principle of term dominance can significantly simplify the assessment process, and the choice of 1n3\frac{1}{n^3} is guided by the notion that higher powers in the denominator control the series' convergence behavior.

The limit comparison test ultimately involves calculating the limit of the ratio of the function of the given series terms to the analogous well-known series terms as n approaches infinity. If this limit is finite and positive, both series either converge or diverge together. Thus, knowing the properties of the 1n3\frac{1}{n^3} series, which is known to converge (as it is a p-series with p > 1), allows us to infer the convergence of the original series. This approach is pivotal in effectively tackling series where direct convergence tests seem cumbersome to apply.

Posted by Gregory 32 minutes ago

Related Problems

Using the comparison test, determine if the series n=112n+1\displaystyle \sum_{n=1}^{\infty} \frac{1}{2^n + 1} is convergent by comparing it to the series n=112n\displaystyle \sum_{n=1}^{\infty} \frac{1}{2^n}.

Determine whether the series n=152n2+4n+3\displaystyle \sum_{n=1}^{\infty} \frac{5}{2n^2 + 4n + 3} is convergent using the comparison test.

Use the limit comparison test to see if the series 1n2+2\frac{1}{\sqrt{n^2 + 2}} converges or diverges.

Determine if the series 13n+5\frac{1}{3^n + 5} converges or diverges using the limit comparison test.