10/2 Limit Comparsion test

The session begins with a recap of the Integral Test and an example to solidify understanding.

The Integral Test states that if $f(x)$ is a positive continuous and decreasing function for $x \geq n$ , then the series $\sum{k=n}^{\infty} f(k)$ converges if and only if the integral $\int{n}^{\infty} f(x) \, dx$ converges.
In the current example, we redefine the series as an integral by letting $f(x) = \frac{1}{x^3}.$

1. Continuous: The function $f(x)$ is continuous because it is defined for $x > 0$ .
2. Positive: For all $x \geq 1$ , $f(x)$ is positive since $\frac{1}{x^3} > 0$ .
3. Decreasing: We show that $f(x)$ is decreasing by calculating the derivative.
- The derivative is:
  $f'(x) = -\frac{3}{x^4},$ which is negative for $x > 0$ , confirming that $f(x)$ is decreasing on $[1, \infty)$ .

With the criteria met, we apply the Integral Test:
- The improper integral to evaluate:
  $\int_{1}^{\infty} \frac{1}{x^3} \, dx$

First, we convert it to a definite integral with a limit:
$\lim{b \to \infty} \int{1}^{b} \frac{1}{x^3} \, dx$
The integral itself:
- The antiderivative of $\frac{1}{x^3}$ is:
  $-\frac{1}{2x^2} + C,$ therefore
  $\int \frac{1}{x^3} \, dx = -\frac{1}{2x^2}$
Evaluating the definite integral: $\lim{b \to \infty} \left[-\frac{1}{2x^2}\right]{1}^{b} = \lim_{b \to \infty} \Bigg(-\frac{1}{2b^2} + \frac{1}{2}\Bigg)$
- As $b$ approaches infinity, $-\frac{1}{2b^2}$ approaches $0$ , thus the integral evaluates to $\frac{1}{2}$ .
Conclusion: Since the integral converges to $\frac{1}{2}$ , by the Integral Test, the series $\sum_{k=1}^{\infty} \frac{1}{k^3}$ converges.

The result of the example illustrates the p-Series Test, which states:
- The series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges if $p > 1$ and diverges if $p \leq 1.$
In the example, $p = 3$ , thus confirming convergence as anticipated.

The follow-up concept introduced is the Limit Comparison Test used to compare two series in terms of convergence or divergence:
- If $an$ and $bn$ are positive, then:
- If $\lim{n \to \infty} \frac{an}{b_n} = L$ where $0 < L < \infty$ , then both series either converge or diverge together.

Consider the series $\sum_{n=1}^{\infty} \left(5n^6 - 3/n^{10} + 12\right)$ and determine its convergence.
Dominating terms for large $n$ dictate the behavior:
- Last terms can be ignored for large $n$ , leading to:
- Essentially compare $5n^6$ with $n^{10}$ to get: $\frac{5n^6}{n^{10}} \sim \frac{5}{n^4}$
Convergence of $\sum \frac{5}{n^4}$ confirmed by p-Series Test since $p = 4 > 1$ .
Therefore, concluding the original series converges as well.

Another example considered: $\sum_{n=1}^{\infty} \frac{\sqrt{n + 1}}{4n - 10}$
Objective is to determine the behavior:
- Behavior at infinity:
- $\sqrt{n + 1} \sim \sqrt{n}$ and $4n \sim 4n$ leading to comparisons of $\frac{\sqrt{n}}{4n}$
- This ratio simplifies to $\frac{1}{4\sqrt{n}}$
Series tested against the divergence of $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ , which also diverges since $p = 1/2 < 1$ .
Consequently, original series diverges too by Limit Comparison Test.

It’s essential to closely observe the dominant terms, especially as $n$ approaches infinity, to determine convergence behavior.
Identifying appropriate comparisons for applying the Limit Comparison Test may take some familiarity but is effectively a useful technique for analyzing series.

Acknowledge the importance of recognizing previous patterns and relationships between series when examining convergence through tests discussed. The concepts learned, such as the Integral Test, the p-Series Test, and the Limit Comparison Test, are critical for thorough understanding of series convergence behavior in calculus.