How to Determine Series Convergence: An Application of the P-Series and Limit Comparison Tests

Understanding the convergence or divergence of infinite series is a crucial aspect of mathematical analysis. In this article, we delve into the process of determining whether a specific series converges or diverges, using the concept of a P-series and the Limit Comparison Test. Let's explore the given series and apply these techniques step-by-step.

Given Series

We are given the series:
$$sum_{n1}^{infty} frac{n^5}{sqrt[3]{n^7n^2}} sum_{n1}^{infty} frac{2}{n^{4/3}}$$

This can be simplified as:

First, let's simplify the fraction: $frac{n^5}{sqrt[3]{n^7n^2}} frac{n^5}{n^{7/3}n^{2/3}} frac{n^5}{n^{(7 2)/3}} frac{n^5}{n^{9/3}} frac{n^5}{n^3} frac{1}{n^{-2}} cdot n^5 frac{2}{n^{4/3}}$ This can be further simplified to: $frac{2}{n^{4/3}}$

Since we have simplified the series to a form that clearly shows a P-series, we can proceed to determine its convergence or divergence.

Convergence through P-Series

A P-series is defined as:
$$sum_{n1}^{infty} frac{1}{n^p}$$

The series converges if and only if (p > 1). In our case, we see that:

$$sum_{n1}^{infty} frac{1}{n^{4/3}}$$

Here, (p frac{4}{3}), which is greater than 1. Hence, by the P-series test, the series:

$$sum_{n1}^{infty} frac{1}{n^{4/3}}$$

converges.

Since we have simplified the given series to:

$$sum_{n1}^{infty} frac{2}{n^{4/3}}$$

which is a constant multiple of a convergent P-series, it follows that:

$$sum_{n1}^{infty} frac{2}{n^{4/3}}$$

also converges.

Using the Limit Comparison Test

To further verify the convergence, we can employ the Limit Comparison Test, which is applicable for positive series. The test states that if:

$$lim_{n rightarrow infty} frac{a_n}{b_n} eq 0, infty$$

then either both series converge or both diverge together.

In our case, let:

$$a_n frac{n^5}{sqrt[3]{n^7n^2}} frac{2}{n^{4/3}}$$

Select a comparison series:

$$b_n frac{n}{n^{7/3}} frac{1}{n^{4/3}}$$

Now, calculate the limit:

$$lim_{n rightarrow infty} frac{a_n}{b_n} lim_{n rightarrow infty} frac{frac{2}{n^{4/3}}}{frac{1}{n^{4/3}}} lim_{n rightarrow infty} 2 2 eq 0, infty$$

This confirms that both series behave similarly and since:

$$sum_{n1}^{infty} frac{1}{n^{4/3}}$$

converges, it follows that:

$$sum_{n1}^{infty} frac{2}{n^{4/3}}$$

also converges.

Conclusion

Both methods, the P-series test and the Limit Comparison Test, confirm that the given series:

$$sum_{n1}^{infty} frac{n^5}{sqrt[3]{n^7n^2}} sum_{n1}^{infty} frac{2}{n^{4/3}}$$

converges.

Related Keywords

P-Series, Limit Comparison Test, Convergence