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Summary of Series Convergence and Divergence Tests

This video provides a comprehensive review of various tests used to determine whether an infinite series converges or diverges.

1. Divergence Test

• Purpose: To quickly identify series that diverge.
• Procedure: Calculate the limit of the sequence $a_n$ as $n$ approaches infinity: $\lim_{n \to \infty} a_n$.
• Conclusion:
  • If $\lim_{n \to \infty} a_n \neq 0$, the series diverges.
  • If $\lim_{n \to \infty} a_n = 0$, the test is inconclusive; another test is needed.

2. Geometric Series

• Form: A constant $a$ multiplied by a common ratio $r$ raised to the power of $n$ or $n-1$: $a \cdot r^n$ or $a \cdot r^{n-1}$.
• Key: Identify the common ratio $r$.
• Conclusion:
  • If $|r| < 1$, the series converges.
  • If $|r| \geq 1$, the series diverges.

3. P-Series Test

• Form: $\frac{1}{n^p}$
• Key: Identify the value of $p$.
• Conclusion:
  • If $p > 1$, the series converges.
  • If $p \leq 1$, the series diverges.

4. Telescoping Series

• Characteristic: Terms in the series cancel each other out when written out.
  • Example: $(1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots$
• Procedure:
  1. Find a formula for the partial sum, $S_n$. This often simplifies to a form like $1 - a_n$ or $1 + a_n$.
  2. Evaluate the limit of the partial sum as $n$ approaches infinity: $\lim_{n \to \infty} S_n$.
• Conclusion:
  • If the limit is a finite value ($L$), the series converges to $L$.
  • If the limit is $\pm \infty$ or does not exist, the series diverges.
• Note: May require partial fraction decomposition to split a single fraction into two that can telescope.

5. Integral Test

• Conditions: For a sequence $a_n$, let $f(x)$ be a function such that $a_n = f(n)$. The function $f(x)$ must be:
  • Positive
  • Continuous
  • Decreasing (for $x \geq N$ for some $N$)
• Procedure: Evaluate the improper integral $\int_{N}^{\infty} f(x) dx$.
• Conclusion:
  • If the integral converges (has a finite value), the original series converges.
  • If the integral diverges ($\pm \infty$ or does not exist), the original series diverges.

6. Ratio Test

• Procedure: Calculate the limit: $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$.
• Conclusion:
  • If $L < 1$, the series converges.
  • If $L > 1$ or $L = \infty$, the series diverges.
  • If $L = 1$, the test is inconclusive.

7. Root Test

• Procedure: Calculate the limit: $L = \lim_{n \to \infty} \sqrt[n]{|a_n|}$.
• Conclusion:
  • If $L < 1$, the series converges.
  • If $L > 1$ or $L = \infty$, the series diverges.
  • If $L = 1$, the test is inconclusive.

8. Direct Comparison Test

• Conditions: Requires two series, $a_n$ and $b_n$. Let $b_n$ be the "larger" series and $a_n$ be the "smaller" series.
• Logic:
  • If the larger series ($\sum b_n$) converges, then the smaller series ($\sum a_n$) also converges.
  • If the smaller series ($\sum a_n$) diverges, then the larger series ($\sum b_n$) also diverges.

9. Limit Comparison Test

• Procedure: Calculate the limit: $L = \lim_{n \to \infty} \frac{a_n}{b_n}$.
• Conditions: Requires two series, $a_n$ and $b_n$. The limit $L$ must be a finite, positive number ($L > 0$).
• Conclusion: If $L$ is a finite positive number, then both series $\sum a_n$ and $\sum b_n$ either converge or diverge together. This test is useful for determining the behavior of one series if the behavior of a similar, simpler series is known.

10. Alternating Series Test

• Form: Series with alternating signs, typically of the form $\sum (-1)^n b_n$ or $\sum (-1)^{n+1} b_n$, where $b_n > 0$.
• Conditions: Two conditions must be met for convergence:
  1. The series must pass the Divergence Test: $\lim_{n \to \infty} b_n = 0$.
  2. The sequence $b_n$ must be decreasing: $b_n \geq b_{n+1}$ for all $n \geq N$ for some $N$.
• Conclusion: If both conditions are met, the alternating series converges.

Absolute vs. Conditional Convergence

• Absolute Convergence: If the series $\sum |a_n|$ converges, then the original series $\sum a_n$ also converges. The series is called absolutely convergent.
• Conditional Convergence: If $\sum |a_n|$ diverges, but the original series $\sum a_n$ converges (e.g., by the Alternating Series Test), then the series is called conditionally convergent.
• Divergence: If $\sum |a_n|$ diverges and $\sum a_n$ also diverges, the series is simply divergent.

Examples and Applications

• Example 1 (Divergence Test): $\sum_{n=1}^{\infty} \frac{2n^2 + 5}{7n^2 - 4}$

  • $\lim_{n \to \infty} \frac{2n^2 + 5}{7n^2 - 4} = \frac{2}{7} \neq 0$.
  • Conclusion: Diverges by the Divergence Test.

• Example 2 (P-Series): $\sum_{n=1}^{\infty} \frac{\sqrt[3]{n}}{n^5}$

  • Simplify to $\sum_{n=1}^{\infty} \frac{1}{n^{14/3}}$. Here $p = \frac{14}{3}$.
  • Since $p > 1$, the series converges by the P-Series Test.

• Example 3 (Geometric Series): $\sum_{n=1}^{\infty} 5 \left(\frac{1}{4}\right)^{n-1}$

  • This is a geometric series with $a=5$ and $r=\frac{1}{4}$.
  • Since $|r| = \frac{1}{4} < 1$, the series converges.
  • The sum is $\frac{a}{1-r} = \frac{5}{1 - 1/4} = \frac{5}{3/4} = \frac{20}{3}$.

• Example 4 (Alternating Series Test): $\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}$

  • Let $b_n = \frac{1}{\sqrt{n}}$.
  • $\lim_{n \to \infty} b_n = \lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0$. (Passes condition 1)
  • $b_n = \frac{1}{\sqrt{n}} > \frac{1}{\sqrt{n+1}} = b_{n+1}$. (Passes condition 2)
  • Conclusion: Converges by the Alternating Series Test.
  • Absolute Convergence Check: $\sum_{n=1}^{\infty} |a_n| = \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{1}{n^{1/2}}$. This is a P-series with $p = \frac{1}{2} \leq 1$, so it diverges.
  • Final Conclusion: The original series is conditionally convergent.

• Example 5 (Telescoping Series): $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$

  • Use partial fractions: $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.
  • The partial sum is $S_n = 1 - \frac{1}{n+1}$.
  • $\lim_{n \to \infty} S_n = \lim_{n \to \infty} (1 - \frac{1}{n+1}) = 1 - 0 = 1$.
  • Conclusion: Converges to 1.

• Example 6 (Direct Comparison Test): $\sum_{n=1}^{\infty} \frac{1}{n^2 + 4}$

  • Compare with $\sum_{n=1}^{\infty} \frac{1}{n^2}$ (a convergent P-series, $p=2>1$).
  • Since $\frac{1}{n^2 + 4} < \frac{1}{n^2}$ for all $n$, and the larger series converges, the smaller series converges by the Direct Comparison Test.

• Example 7 (Integral Test): $\sum_{n=3}^{\infty} \frac{1}{\sqrt{n-2}}$

  • Let $f(x) = \frac{1}{\sqrt{x-2}}$. This function is positive, continuous, and decreasing for $x \geq 3$.
  • Evaluate $\int_{3}^{\infty} \frac{1}{\sqrt{x-2}} dx$. Using u-substitution ($u=x-2$), the integral becomes $\int_{1}^{\infty} u^{-1/2} du = [2\sqrt{u}]{1}^{\infty} = [2\sqrt{x-2}]{3}^{\infty}$.
  • $\lim_{a \to \infty} [2\sqrt{x-2}]{3}^{a} = \lim{a \to \infty} (2\sqrt{a-2} - 2\sqrt{1}) = \infty$.
  • Conclusion: The integral diverges, so the series diverges by the Integral Test.

• Example 8 (Limit Comparison Test): $\sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^3 + 2}$

  • Compare with $\sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^3} = \sum_{n=1}^{\infty} \frac{1}{n^{2.5}}$. This is a convergent P-series ($p=2.5>1$).
  • Calculate $L = \lim_{n \to \infty} \frac{\sqrt{n}/(n^3+2)}{1/n^{2.5}} = \lim_{n \to \infty} \frac{n^{2.5}\sqrt{n}}{n^3+2} = \lim_{n \to \infty} \frac{n^3}{n^3+2} = 1$.
  • Since $L=1$ (finite and positive), and the comparison series converges, the original series converges by the Limit Comparison Test.

• Example 9 (Root Test): $\sum_{n=1}^{\infty} \left(\frac{3n^2 - 9}{7n^2 + 4}\right)^n$

  • Calculate $L = \lim_{n \to \infty} \sqrt[n]{\left|\left(\frac{3n^2 - 9}{7n^2 + 4}\right)^n\right|} = \lim_{n \to \infty} \left|\frac{3n^2 - 9}{7n^2 + 4}\right|$.
  • Using L'Hopital's Rule twice or by comparing degrees, the limit is $\frac{3}{7}$.
  • Since $L = \frac{3}{7} < 1$, the series converges by the Root Test.

• Example 10 (Ratio Test): $\sum_{n=1}^{\infty} \frac{2^n}{n!}$

  • $a_n = \frac{2^n}{n!}$, $a_{n+1} = \frac{2^{n+1}}{(n+1)!}$.
  • Calculate $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{2^{n+1}}{(n+1)!} \cdot \frac{n!}{2^n} \right| = \lim_{n \to \infty} \left| \frac{2}{n+1} \right|$.
  • $L = 0$.
  • Since $L = 0 < 1$, the series converges by the Ratio Test.