July 01, 2026 - Calculus 2 - Comparison and Alternating Series Tests

Review of the P-Series Test

• The lecture begins with a recap of the P-Series Test, which was proved using the integral test.

• The P-Series is defined as the series: $\sum_{n=1}^{\infty} \frac{1}{n^p}$

• It converges if p > 1.

• It diverges otherwise (if $p \leq 1$).

• The P-series is a primary tool used as a benchmark for the comparison tests.

The Direct Comparison Test

• The Direct Comparison Test involves comparing a given sequence $a_n$ to a known sequence $b_n$.

• Requirement: The inequality must hold for some capital letter $N$. This means the sequences can behave differently at the beginning, but after a certain point $n \geq N$, the relationship must hold true.

• Part 1: Convergence

  • If $a_n \leq b_n$ for all $n \geq N$.

  • If the series $\sum b_n$ converges, then the series $\sum a_n$ also converges.

  • Intuition: If the sum of the larger sequence is a finite number, then the sum of a sequence that is smaller must also be a finite number.

• Part 2: Divergence

  • If $a_n \geq b_n$ for all $n \geq N$.

  • If the series $\sum b_n$ diverges (sum equals infinity), then the series $\sum a_n$ also diverges.

  • Intuition: If you add up an infinite amount of smaller numbers and get infinity, adding up larger numbers will also result in infinity (e.g., if adding 1s infinitely is infinity, adding 2s infinitely is also infinity).

• Strategy for choosing a comparison series:

  • If you suspect the series diverges, choose a comparison series $b_n$ that is smaller than $a_n$ and is known to diverge.

  • If you suspect the series converges, choose a comparison series $b_n$ that is larger than $a_n$ and is known to converge.

Examples of the Direct Comparison Test

• Example 1: Rational function with a shifted denominator

  • Series: $\sum \frac{1}{n - \frac{1}{2}}$

  • Choice of comparison: The Harmonic Series ($\sum \frac{1}{n}$), which is a P-series with $p = 1$. The harmonic series diverges.

  • Verification:

  • $n = 1 \rightarrow a_1 = \frac{1}{1 - 0.5} = 2$ vs. $b_1 = 1$

  • $n = 2 \rightarrow a_2 = \frac{1}{2 - 0.5} = \frac{2}{3}$ vs. $b_2 = 0.5$

  • Since \frac{1}{n - \frac{1}{2}} > \frac{1}{n} , the original series is larger than a divergent series.

  • Conclusion: The series diverges by the comparison test.

• Example 2: Higher degree polynomial

  • Series: $\sum \frac{1}{n^3 + 1}$

  • Choice of comparison: $\sum \frac{1}{n^3}$. This is a P-series with p = 3 > 1, so it converges.

  • Verification: Show that $\frac{1}{n^3 + 1} \leq \frac{1}{n^3}$.

  • Cross multiplying: $n^3 \leq n^3 + 1$ (clearly true).

  • Conclusion: The series converges by the comparison test.

• Example 3: Exponential denominator

  • Series: $\sum \frac{1}{2^n + 1}$

  • Choice of comparison: If the "+1" is ignored, it looks like a Geometric Series: $\sum \frac{1}{2^n}$ with $r = \frac{1}{2}$. Since -1 < r < 1, it converges.

  • Verification: Show that $\frac{1}{2^n + 1} \leq \frac{1}{2^n}$.

  • Cross multiplying: $2^n \leq 2^n + 1$ (clearly true).

  • Conclusion: The series converges by the comparison test.

• Example 4: Natural Logarithm

  • Series: $\sum \frac{1}{\ln(n)}$

  • Choice of comparison: The Harmonic series $\sum \frac{1}{n}$ (divergent).

  • Verification: Check if $\ln(n) \leq n$.

  • Using e: $e^{\ln(n)} \leq e^n$ implies $n \leq e^n$, which is true for $n \geq 1$.

  • Therefore, \frac{1}{\ln(n)} > \frac{1}{n} .

  • Conclusion: Since it is larger than a divergent series, it also diverges.

Limit Comparison Test (LCT)

• The Limit Comparison Test is useful when direct inequalities are difficult to establish (e.g., when terms are subtracted in the denominator).

• Requirement: Both series $a_n$ and $b_n$ must have positive terms.

• Formula: Evaluate the limit $L = \lim_{n \rightarrow \infty} \frac{a_n}{b_n}$.

• Three Cases of the Limit Comparison Test:

  1. If 0 < L < \infty: Both series behave the same way (both converge or both diverge). This occurs when the terms grow at the same "speed."

  2. If $L = 0$: The denominator $b_n$ grows faster than the numerator $a_n$. If $\sum b_n$ converges, then $\sum a_n$ also converges.

  3. If $L = \infty$: The numerator $a_n$ grows faster than the denominator $b_n$. If $\sum b_n$ diverges, then $\sum a_n$ also diverges.

Examples of the Limit Comparison Test

• Example 1: Square root denominator

  • Series: $\sum \frac{1}{\sqrt{n} + 1}$

  • Compare to: $b_n = \frac{1}{\sqrt{n}} = \frac{1}{n^{1/2}}$ (P-series with $p = 0.5$, diverges).

  • Limit: $\lim_{n \rightarrow \infty} \frac{\frac{1}{\sqrt{n} + 1}}{\frac{1}{\sqrt{n}}} = \lim_{n \rightarrow \infty} \frac{\sqrt{n}}{\sqrt{n} + 1}$.

  • Using L'Hôpital's Rule: $\lim_{n \rightarrow \infty} \frac{\frac{1}{2\sqrt{n}}}{\frac{1}{2\sqrt{n}}} = 1$.

  • Conclusion: Since $L=1$ and $b_n$ diverges, the original series diverges.

• Example 2: Geometric vs. shifted geometric

  • Series: $\sum \frac{2^n + 1}{3^n + 1}$

  • Compare to: $b_n = \frac{2^n}{3^n} = (\frac{2}{3})^n$ (Geometric, r = \frac{2}{3} < 1, converges).

  • Calculation: Using L'Hôpital's or splitting the fraction shows the limit is 1.

  • Conclusion: The series converges.

• Example 3: Logarithm and P-series (The "In-Between" Strategy)

  • Series: $\sum \frac{\ln(n)}{n^2}$

  • Attempting $\sum \frac{1}{n}$ (diverges): Limit is 0. Case 2 check: Divergence doesn't tell us anything if limit is 0 (Inconc lusive).

  • Attempting $\sum \frac{1}{n^2}$ (converges): Limit is infinity. Case 3 check: Convergence doesn't tell us anything if limit is infinity (Inconclusive).

  • Correct Choice: Select a series "in between" such as $b_n = \frac{1}{n^{1.5}} = \frac{1}{n^{3/2}}$ (convergent P-series).

  • Calculation: The limit is 0. Since the comparison series converges and the limit is 0, the numerator series also converges.

The Alternating Series Test (AST)

• Definition: An alternating series takes the form $\sum (-1)^n a_n$ or $\sum (-1)^{n+1} a_n$, where terms alternate between positive and negative.

• Telescopic Series: Some alternating series are telescopic, where middle terms cancel out, though most discussed here are not.

• The Test: An alternating series converges if it satisfies two conditions:

  1. The sequence of terms $a_n$ is decreasing: $a_{n+1} \leq a_n$ for all $n$.

  2. The limit of the terms is zero: $\lim_{n \rightarrow \infty} a_n = 0$.

• Note: The Alternating Series Test does not directly prove divergence; if it fails, one should use the Divergence Test (if $\lim_{n \rightarrow \infty} a_n \neq 0$, the series diverges).

Examples of the Alternating Series Test

• Example 1: Alternating P-series

  • Series: $\sum \frac{(-1)^n}{n^2}$

  • Term $a_n = \frac{1}{n^2}$.

  • Decreasing: $\frac{1}{(n+1)^2} \leq \frac{1}{n^2}$ is true because (n+1)^2 > n^2 .

  • Limit: $\lim_{n \rightarrow \infty} \frac{1}{n^2} = 0$.

  • Conclusion: Converges.

• Example 2: Failing AST

  • Series: $\sum (-1)^n \frac{n}{n+1}$

  • Limit Check: $\lim_{n \rightarrow \infty} \frac{n}{n+1} = 1$.

  • Result: Fails condition 2 of AST. By the Divergence Test, the limit of the terms does not go to zero, so the series diverges.

• Example 3: Logarithms in Alternating Series

  • Series: $\sum (-1)^n \frac{\ln(n)}{2^n}$

  • Limit: Using L'Hôpital's Rule: $\lim_{n \rightarrow \infty} \frac{\frac{1}{n}}{2^n \ln(2)} = 0$. Condition 2 satisfied.

  • Decreasing: Cross multiplication of terms shows \ln(n+1) \cdot 2^n < \ln(n) \cdot 2^{n+1} for large $n$. Condition 1 satisfied.

  • Conclusion: Converges.

Questions & Discussion

• Question: Why use $n$ or $C$ in the notation?

• Answer: $n$ is used when talking about integers (discrete sequences); $C$ would refer to any real number (continuous functions).

• Question: If it is not a P-series, how do we find a comparison?

• Answer: For rational functions, look at the highest degree of the polynomial. For exponential functions, look at the base to determine a geometric series comparison. Sometimes it is trial and error.

• Question: What about the orientation mentioned?

• Answer: There is an FSU orientation happening tomorrow, so the lecture notes or test will be posted online for those who miss it.

• Question: What is a telescopic series?

• Answer: It is a series where terms in the middle cancel, similar to collapsing a telescope. The lecturer offered to show an example after class for the student who missed it.

• Question: Is the Taylor and Maclaurin section a separate module?

• Answer: Yes, power series (including Taylor and Maclaurin) will be the focus of the final two weeks of the course.

Administrative Notes

• Take-home Test: A take-home test covering all series tests (P-test, Comparison, Limit Comparison, Alternating) will be given out this Thursday.

• Schedule: The test should be printed out and returned on Monday.

• Upcoming Topics: Two more convergence tests will be covered in the next session before moving into Power Series.