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Determining Convergence or Divergence of Series: A Step-by-Step Guide

February 05, 2025Technology1326
Determining Convergence or Divergence of Series: A Step-by-Step Guide

Determining Convergence or Divergence of Series: A Step-by-Step Guide

When dealing with series in mathematics, one crucial aspect is determining whether a given series converges or diverges. This article focuses on the series (sum_{n1}^{infty} frac{5sqrt{n} 1}{sqrt{n^3 - 2n^2 - 3}}).

Introduction to Convergence and Divergence

Convergence refers to a series where the sum of the terms approaches a finite limit as the number of terms increases. In contrast, divergence occurs when the sum of the series does not approach a finite limit. Many tests are available to determine convergence or divergence, such as the limit comparison test, nth-term test, and integral test.

Applying the Limit Comparison Test

To analyze the series (sum_{n1}^{infty} frac{5sqrt{n} 1}{sqrt{n^3 - 2n^2 - 3}}), we start by simplifying the general term for large n. Let's rewrite the general term:

(frac{5sqrt{n} 1}{sqrt{n^3 - 2n^2 - 3}} frac{sqrt{n} (5 frac{1}{sqrt{n}})}{sqrt{n^3 (1 - frac{2}{n} - frac{3}{n^3})}} frac{5 frac{1}{sqrt{n}}}{n sqrt{1 - frac{2}{n} - frac{3}{n^3}}})

For large n, the term (frac{1}{sqrt{n}}) and (frac{3}{n^3}) approach zero, making the denominator approximately (nsqrt{1}), which simplifies to (n).

Asymptotic Behavior and Comparison with the Harmonic Series

The simplified term (frac{5 frac{1}{sqrt{n}}}{n}) is asymptotic to (frac{5}{n}) as n grows large. The harmonic series, (sum_{n1}^{infty} frac{1}{n}), is known to diverge. Comparing our series to the harmonic series using the limit comparison test can provide insight into the behavior of our given series.

Limit Comparison Test Calculation

To apply the limit comparison test, we compute the limit:

(L lim_{n to infty} frac{frac{5sqrt{n} 1}{sqrt{n^3 - 2n^2 - 3}}}{frac{1}{n}})

Substituting and simplifying:

(L lim_{n to infty} frac{n (5sqrt{n} 1)}{sqrt{n^3 - 2n^2 - 3}} lim_{n to infty} frac{5n^{3/2} n}{nsqrt{1 - frac{2}{n} - frac{3}{n^3}}} lim_{n to infty} frac{5 frac{1}{sqrt{n}}}{sqrt{1 - frac{2}{n} - frac{3}{n^3}}})

As n grows large, the limit approaches 5, which is a finite and non-zero value. Thus, by the limit comparison test, the series (sum_{n1}^{infty} frac{5sqrt{n} 1}{sqrt{n^3 - 2n^2 - 3}}) diverges because the harmonic series diverges.

Conclusion

The limit comparison test has shown that the given series (sum_{n1}^{infty} frac{5sqrt{n} 1}{sqrt{n^3 - 2n^2 - 3}}) diverges. Since the terms do not approach zero as n grows, another common test (the nth-term test) could also provide this conclusion.

Key Test Methods

Limit Comparison Test: Comparing the given series to another series known to converge or diverge. Hospital's Rule: For more complex series that cannot be simplified easily. (Not necessary in this case as the limit comparison test suffices). Nth-Term Test: (Necessary Conditions for Convergence) The terms of the series must approach zero for convergence.

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