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

Step 2: Create inequality

$0 < \frac{2 + \cos n}{n^3} \leq \frac{3}{n^3}$

Step 3: Check comparison series

$\sum_{n=1}^{\infty} \frac{3}{n^3} = 3\sum_{n=1}^{\infty} \frac{1}{n^3}$

This is a p-series with $p = 3 > 1$, so it converges.

Step 4: Apply DCT

Since $0 < \frac{2+\cos n}{n^3} \leq \frac{3}{n^3}$ and $\sum \frac{3}{n^3}$ converges:

By Direct Comparison Test, the series converges.

Compare to $b_n = \frac{1}{n \ln n}$

Step 2: What about $\sum \frac{1}{n \ln n}$?

From Integral Test (previous topic), we know:

$\sum_{n=2}^{\infty} \frac{1}{n \ln n} \text{ diverges}$

(The integral $\int \frac{1}{x \ln x}dx = \ln|\ln x| \to \infty$)

Step 3: Use Limit Comparison Test

$\lim_{n \to \infty} \frac{\frac{1}{n\ln n - 1}}{\frac{1}{n\ln n}} = \lim_{n \to \infty} \frac{n \ln n}{n \ln n - 1}$

Divide numerator and denominator by $n \ln n$:

$= \lim_{n \to \infty} \frac{1}{1 - \frac{1}{n\ln n}} = \frac{1}{1-0} = 1$

Step 4: Apply LCT

Since $L = 1 > 0$ and $\sum \frac{1}{n \ln n}$ diverges:

By LCT, the series diverges.