Question

Use the special comparison test to find whether the following series converge or diverge. $$\sum_{n=3}^{\infty} \frac{(n-\ln n)^{2}}{5 n^{4}-3 n^{2}+1}$$

Short answer

The series converges by the Limit Comparison Test.

Step-by-step solution

- Formulate the Comparison

Identify a simpler series to compare with. The series under consideration is \(\frac{(n-\text{ln} n)^{2}}{5 n^{4}-3 n^{2}+1}\). Notice that for larger values of n, \(5 n^{4}\) dominates the denominator. So, compare it with \(b_n = \frac{n^2}{5 n^4} = \frac{1}{5 n^2}\).

- Write the Comparison Series

The comparison series is \(\frac{1}{5 n^{2}}\). Recognize this as a p-series \(\frac{1}{n^p}\) with \(p = 2\).

- Determine Convergence of the Comparison Series

A p-series \(\frac{1}{n^p}\) converges if \(p > 1\). Here, \(\frac{1}{n^2}\) converges because \(p = 2 > 1\). Therefore, \(\frac{1}{5 n^2}\) also converges.

- Apply the Limit Comparison Test

Compute the limit\ \(\text{lim}_{n \to \infty} \frac{a_n}{b_n} = \text{lim}_{n \to \infty} \frac{\frac{(n - \text{ln} n)^2}{5 n^4 - 3 n^2 + 1}}{\frac{1}{5 n^2}} = \text{lim}_{n \to \infty} \frac{5 n^2 (n - \text{ln} n)^2}{5 n^4 - 3 n^2 + 1} \).

- Simplify the Limit

Simplify the expression inside the limit: \(\text{lim}_{n \to \infty} \frac{5 n^2 (n - \text{ln} n)^2}{5 n^4 - 3 n^2 + 1}\).For large n: divide by \^4\ to get \(\text{lim}_{n \to \infty} \frac{5 (1 - \frac{\text{ln} n}{n})^2}{5 - \frac{3}{n^2} + \frac{1}{n^4}} = \text{lim}_{n \to \infty} \frac{5 (1 - 0)^2}{5 - 0 + 0} = 1\).

- Conclude the Convergence

Since the limit is a finite positive number (\text{L} = 1), and we know that the comparison series \(\frac{1}{5 n^2}\) converges, by the Limit Comparison Test, \(\frac{(n - \text{ln} n)^2}{5 n^4 - 3 n^2 + 1}\) also converges.

Key concepts

Special Comparison Test

The Special Comparison Test helps us determine if a series converges or diverges by comparing it to a simpler, well-understood series. To use this effectively, we should:

Identify a series that resembles the given series for large values of the index (n).
Write down both the given series and the comparison series.
Use properties of the comparison series to draw conclusions about the given series.

For our exercise, we compare the given series \(\frac{(n-\text{ln} n)^{2}}{5 n^{4}-3 n^{2}+1}\) with the simpler series \(\frac{1}{5 n^{2}}\), which is easier to analyze. We find that the chosen comparison series is a p-series. Let's talk about p-series next.

P-Series

A p-series takes the form \(\frac{1}{n^p}\), where p is a constant. The convergence of a p-series depends entirely on the value of p:

If \( p > 1 \), the p-series converges.
If \( p \leq 1 \), the p-series diverges.

In our example, the comparison series is \(\frac{1}{5 n^2}\). Here, we have \( p = 2 \), which means it converges. This insight provides a solid foundation to proceed with more specific tests, like the Limit Comparison Test.

Limit Comparison Test

The Limit Comparison Test helps us confirm conclusions drawn from comparing two series. It's useful when direct comparison isn't straightforward. To apply this test:

Compute the limit \(\text{lim}_{n \to \text{infty}} \frac{a_n}{b_n}\), where \ a_n \ is the given series and \ b_n \ is the comparison series.
If the limit is a finite positive number \ (L > 0) \, both series either converge or diverge together.

For our series, calculate \ \text{lim}_{n \to \text{infty}} \frac{5 n^2 (n - \text{ln} n)^2}{5 n^4 - 3 n^2 + 1} \ and simplify. By dividing common factors, the limit becomes \ \text{lim}_{n \to \text{infty}} \frac{5 (1 - \frac{\text{ln} n}{n})^2}{5 - \frac{3}{n^2} + \frac{1}{n^4}} = 1 \.
Since this limit is a finite positive number, it confirms that the given series converges just like the comparison series.

Convergence

Convergence in the context of series means that as we sum more and more terms, the total approaches a fixed number. Mathematically, a series \sum a_n \ converges if the sequence of its partial sums has a finite limit as \( n \to \text{infty} \).

To determine convergence, we use tests like the Special Comparison Test and the Limit Comparison Test. For our exercise:
1. Identify \( a_n = \frac{(n-\text{ln} n)^{2}}{5 n^{4}-3 n^{2}+1} \)
2. Compare it with \( b_n = \frac{1}{5 n^2} \).
3. Use these tests to simplify and confirm that the series converges.

This systematic approach helps us understand whether a series converges or diverges and, more importantly, why.