# 🎯 Comparison Tests ## The Big Idea If you can't evaluate $\sum a_n$ directly, **compare it to a series you know**! > **💡 Key Idea**: If $a_n$ is smaller than a convergent series, it converges. If $a_n$ is larger than a divergent series, it diverges. --- ## Direct Comparison Test (DCT) Suppose $0 \leq a_n \leq b_n$ for all $n \geq N$ (some starting point). ### Part 1: If $\sum b_n$ converges, then $\sum a_n$ converges. **Reason**: $a_n$ is smaller than convergent series, so it must also converge. > **"Smaller than convergent → convergent"** --- ### Part 2: If $\sum a_n$ diverges, then $\sum b_n$ diverges. **Reason**: $b_n$ is larger than divergent series, so it must also diverge. > **"Larger than divergent → divergent"** --- ## How to Use Direct Comparison Test **Step 1**: Find a comparison series $\sum b_n$ that you know converges or diverges - Usually a **p-series** or **geometric series**! **Step 2**: Show $0 \leq a_n \leq b_n$ (or $0 \leq b_n \leq a_n$) **Step 3**: Apply the test --- ## Example 1: Use DCT Determine if $\sum_{n=1}^{\infty} \frac{1}{2^n + n}$ converges. **Step 1: Find comparison** For $n \geq 1$: $2^n + n > 2^n$ So: $\frac{1}{2^n + n} < \frac{1}{2^n}$ Compare to $\sum \frac{1}{2^n}$ (geometric with $r = \frac{1}{2}$). --- **Step 2: Check comparison series** $$\sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{1/2}{1 - 1/2} = 1$$ (converges!) --- **Step 3: Apply DCT** Since $0 < \frac{1}{2^n + n} < \frac{1}{2^n}$ and $\sum \frac{1}{2^n}$ converges: **By DCT, $\sum \frac{1}{2^n + n}$ converges.** --- ## Example 2: Use DCT to Show Divergence Determine if $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} + 5}$ converges. **Step 1: Find comparison** For $n \geq 1$: $\sqrt{n} + 5 < \sqrt{n} + 5\sqrt{n} = 6\sqrt{n}$ (not helpful) Better: $\sqrt{n} + 5 < \sqrt{n} + \sqrt{n} = 2\sqrt{n}$ (only true for large $n$) Actually, for large $n$: $\sqrt{n} + 5 \approx \sqrt{n}$ So: $\frac{1}{\sqrt{n} + 5} \approx \frac{1}{\sqrt{n}}$ But we need an inequality! Try: For $n \geq 1$: $\sqrt{n} + 5 > \sqrt{n}$ So: $\frac{1}{\sqrt{n} + 5} < \frac{1}{\sqrt{n}}$ (wrong direction for divergence!) --- **Better approach**: $\sqrt{n} + 5 < \sqrt{n} + 5n$ (true for $n \geq 1$... no) Actually, for $n \geq 25$: $\sqrt{n} + 5 \leq 2\sqrt{n}$ So: $\frac{1}{\sqrt{n} + 5} \geq \frac{1}{2\sqrt{n}}$ --- **Step 2: Check comparison series** $$\sum \frac{1}{\sqrt{n}} = \sum \frac{1}{n^{1/2}}$$ is a p-series with $p = \frac{1}{2} < 1$ (diverges!) So $\sum \frac{1}{2\sqrt{n}} = \frac{1}{2}\sum \frac{1}{\sqrt{n}}$ also diverges. --- **Step 3: Apply DCT** Since $\frac{1}{\sqrt{n} + 5} \geq \frac{1}{2\sqrt{n}}$ for $n \geq 25$ and $\sum \frac{1}{2\sqrt{n}}$ diverges: **By DCT, $\sum \frac{1}{\sqrt{n} + 5}$ diverges.** --- ## When DCT is Hard to Use Sometimes it's hard to prove the inequality! **Example**: Is $\frac{n}{n^2 + 1}$ less than or greater than $\frac{1}{n}$? $$\frac{n}{n^2 + 1} \text{ vs } \frac{1}{n}$$ Cross-multiply: $n^2$ vs $n^2 + 1$ So $\frac{n}{n^2 + 1} < \frac{1}{n}$ ✓ But if inequality is close or complicated, use **Limit Comparison Test** instead! --- ## Limit Comparison Test (LCT) Let $a_n > 0$ and $b_n > 0$ for all $n$. If: $$\lim_{n \to \infty} \frac{a_n}{b_n} = L$$ where $L$ is finite and $L > 0$, then: **$\sum a_n$ and $\sum b_n$ both converge or both diverge!** > **💡 Key Idea**: If $\frac{a_n}{b_n} \to L > 0$, then $a_n$ and $b_n$ behave the same way! --- ## Special Cases of LCT - If $L = 0$ and $\sum b_n$ converges → $\sum a_n$ converges - If $L = \infty$ and $\sum b_n$ diverges → $\sum a_n$ diverges - If $L = 0$ and $\sum b_n$ diverges → inconclusive - If $L = \infty$ and $\sum b_n$ converges → inconclusive **Best to aim for $0 < L < \infty$!** --- ## How to Use Limit Comparison Test **Step 1**: Choose a comparison series $\sum b_n$ (usually p-series or geometric) **Tip**: Look at the **dominant terms** (highest powers) in numerator and denominator. **Step 2**: Compute $\lim_{n \to \infty} \frac{a_n}{b_n}$ **Step 3**: If $0 < L < \infty$, both series have same convergence behavior --- ## Example 3: Use LCT Determine if $\sum_{n=1}^{\infty} \frac{3n^2 + 5}{n^4 - 2n + 1}$ converges. **Step 1: Find dominant terms** Numerator: $3n^2$ dominates Denominator: $n^4$ dominates So $a_n \approx \frac{3n^2}{n^4} = \frac{3}{n^2}$ **Compare to $b_n = \frac{1}{n^2}$** (p-series with $p = 2 > 1$, converges!) --- **Step 2: Compute limit** $$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{3n^2 + 5}{n^4 - 2n + 1}}{\frac{1}{n^2}}$$ $$= \lim_{n \to \infty} \frac{(3n^2 + 5)n^2}{n^4 - 2n + 1}$$ $$= \lim_{n \to \infty} \frac{3n^4 + 5n^2}{n^4 - 2n + 1}$$ Divide by $n^4$: $$= \lim_{n \to \infty} \frac{3 + \frac{5}{n^2}}{1 - \frac{2}{n^3} + \frac{1}{n^4}} = \frac{3}{1} = 3$$ --- **Step 3: Apply LCT** $L = 3$ (finite and positive), and $\sum \frac{1}{n^2}$ converges. **By LCT, $\sum \frac{3n^2 + 5}{n^4 - 2n + 1}$ converges.** --- ## Example 4: Use LCT Determine if $\sum_{n=1}^{\infty} \frac{\sqrt{n} + 1}{n^2}$ converges. **Step 1: Dominant terms** Numerator: $\sqrt{n} = n^{1/2}$ Denominator: $n^2$ So $a_n \approx \frac{n^{1/2}}{n^2} = \frac{1}{n^{3/2}}$ **Compare to $b_n = \frac{1}{n^{3/2}}$** (p-series with $p = \frac{3}{2} > 1$, converges!) --- **Step 2: Compute limit** $$\lim_{n \to \infty} \frac{\frac{\sqrt{n} + 1}{n^2}}{\frac{1}{n^{3/2}}} = \lim_{n \to \infty} \frac{(\sqrt{n} + 1)n^{3/2}}{n^2}$$ $$= \lim_{n \to \infty} \frac{n^{1/2} \cdot n^{3/2} + n^{3/2}}{n^2}$$ $$= \lim_{n \to \infty} \frac{n^2 + n^{3/2}}{n^2}$$ Divide by $n^2$: $$= \lim_{n \to \infty} \left(1 + \frac{1}{n^{1/2}}\right) = 1 + 0 = 1$$ --- **Step 3: Apply LCT** $L = 1 > 0$ and $\sum \frac{1}{n^{3/2}}$ converges. **By LCT, the series converges.** --- ## Example 5: LCT for Divergence Determine if $\sum_{n=1}^{\infty} \frac{2n + \sin n}{3n - 1}$ converges. **Step 1: Dominant terms** Numerator: $2n$ dominates (sin n is bounded) Denominator: $3n$ dominates So $a_n \approx \frac{2n}{3n} = \frac{2}{3}$ This is like a constant series! **Compare to $b_n = 1$** (or any constant) --- **Actually better**: $a_n \approx \frac{2n}{3n} = \frac{2}{3}$, compare to $b_n = \frac{1}{n^0} = 1$ Wait, we need a series. Since $a_n$ stays around $\frac{2}{3}$ (doesn't go to 0), use **nth term test**! $$\lim_{n \to \infty} \frac{2n + \sin n}{3n - 1} = \lim_{n \to \infty} \frac{2 + \frac{\sin n}{n}}{3 - \frac{1}{n}} = \frac{2}{3} \neq 0$$ **By nth term test: diverges!** --- ## Choosing the Right Comparison **For rational functions** (polynomials): - Keep only highest power terms - $\frac{n^a}{n^b} = \frac{1}{n^{b-a}}$ (compare to p-series with $p = b - a$) --- **For exponentials**: - $2^n, e^n$ grow much faster than polynomials - Compare to geometric series --- **For square roots and powers**: - $\sqrt{n} = n^{1/2}$ - $\sqrt[3]{n} = n^{1/3}$ - Compare to appropriate p-series --- ## DCT vs LCT: Which to Use? **Use DCT when**: - Inequality is obvious - You need rigorous proof **Use LCT when**: - Inequality is hard to prove - You just need to check convergence - Working with complicated expressions > **💡 LCT is usually easier!** Most students prefer it. --- ## ⚠️ Common Mistakes ### Mistake 1: Wrong Comparison Direction **WRONG**: "$a_n < b_n$ and $\sum b_n$ diverges, so $\sum a_n$ diverges" **RIGHT**: Smaller than divergent tells you nothing! Need "larger than divergent" for divergence. --- ### Mistake 2: Limit is 0 or Infinity If $\lim \frac{a_n}{b_n} = 0$ or $\infty$, basic LCT is **inconclusive** (unless special cases apply). Choose a different comparison series! --- ### Mistake 3: Negative Terms Comparison tests require $a_n, b_n \geq 0$! For alternating or negative terms, use different tests. --- ### Mistake 4: Ignoring Smaller Terms **WRONG**: "$\frac{n^2 + 5n + 1}{n^3}$ is like $\frac{n^2}{n^3}$... wait, what about the $5n$?" **RIGHT**: For large $n$, lower powers don't matter! $n^2$ dominates $5n$ and $1$. --- ## 📝 Practice Strategy 1. **Always try nth term test first**: If $\lim a_n \neq 0$, diverges immediately! 2. **Find dominant terms**: Highest powers in numerator and denominator 3. **Choose comparison**: Usually p-series $\sum \frac{1}{n^p}$ or geometric $\sum r^n$ 4. **Prefer LCT**: Easier than proving inequalities 5. **For DCT**: Make sure inequality goes the right direction! 6. **Check your limit**: Should get finite positive number for LCT 7. **Memorize**: $\sum \frac{1}{n^p}$ converges if $p > 1$, diverges if $p \leq 1$

Direct and Limit Comparison Tests