Direct and Limit Comparison Tests

Comparing series to determine convergence

🎯 Comparison Tests

The Big Idea

If you can't evaluate ∑an\sum a_n directly, compare it to a series you know!

💡 Key Idea: If ana_n is smaller than a convergent series, it converges. If ana_n is larger than a divergent series, it diverges.

Direct Comparison Test (DCT)

Suppose 0≤an≤bn0 \leq a_n \leq b_n for all n≥Nn \geq N (some starting point).

Part 1: If ∑bn\sum b_n converges, then ∑an\sum a_n converges.

Reason: ana_n is smaller than convergent series, so it must also converge.

"Smaller than convergent → convergent"

Part 2: If ∑an\sum a_n diverges, then ∑bn\sum b_n diverges.

Reason: bnb_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 ∑bn\sum b_n that you know converges or diverges

  • Usually a p-series or geometric series!

Step 2: Show 0≤an≤bn0 \leq a_n \leq b_n (or 0≤bn≤an0 \leq b_n \leq a_n)

Step 3: Apply the test

Example 1: Use DCT

Determine if ∑n=1∞12n+n\sum_{n=1}^{\infty} \frac{1}{2^n + n} converges.

Step 1: Find comparison

For n≥1n \geq 1: 2n+n>2n2^n + n > 2^n

So: 12n+n<12n\frac{1}{2^n + n} < \frac{1}{2^n}

Compare to ∑12n\sum \frac{1}{2^n} (geometric with r=12r = \frac{1}{2}).

Step 2: Check comparison series

∑n=1∞12n=1/21−1/2=1\sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{1/2}{1 - 1/2} = 1 (converges!)

Step 3: Apply DCT

Since 0<12n+n<12n0 < \frac{1}{2^n + n} < \frac{1}{2^n} and ∑12n\sum \frac{1}{2^n} converges:

By DCT, ∑12n+n\sum \frac{1}{2^n + n} converges.

Example 2: Use DCT to Show Divergence

Determine if ∑n=1∞1n+5\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} + 5} converges.

Step 1: Find comparison

For n≥1n \geq 1: n+5<n+5n=6n\sqrt{n} + 5 < \sqrt{n} + 5\sqrt{n} = 6\sqrt{n} (not helpful)

Better: n+5<n+n=2n\sqrt{n} + 5 < \sqrt{n} + \sqrt{n} = 2\sqrt{n} (only true for large nn)

Actually, for large nn: n+5≈n\sqrt{n} + 5 \approx \sqrt{n}

So: 1n+5≈1n\frac{1}{\sqrt{n} + 5} \approx \frac{1}{\sqrt{n}}

But we need an inequality! Try:

For n≥1n \geq 1: n+5>n\sqrt{n} + 5 > \sqrt{n}

So: 1n+5<1n\frac{1}{\sqrt{n} + 5} < \frac{1}{\sqrt{n}} (wrong direction for divergence!)

Better approach: n+5<n+5n\sqrt{n} + 5 < \sqrt{n} + 5n (true for n≥1n \geq 1... no)

Actually, for n≥25n \geq 25: n+5≤2n\sqrt{n} + 5 \leq 2\sqrt{n}

So: 1n+5≥12n\frac{1}{\sqrt{n} + 5} \geq \frac{1}{2\sqrt{n}}

Step 2: Check comparison series

∑1n=∑1n1/2\sum \frac{1}{\sqrt{n}} = \sum \frac{1}{n^{1/2}} is a p-series with p=12<1p = \frac{1}{2} < 1 (diverges!)

So ∑12n=12∑1n\sum \frac{1}{2\sqrt{n}} = \frac{1}{2}\sum \frac{1}{\sqrt{n}} also diverges.

Step 3: Apply DCT

Since 1n+5≥12n\frac{1}{\sqrt{n} + 5} \geq \frac{1}{2\sqrt{n}} for n≥25n \geq 25 and ∑12n\sum \frac{1}{2\sqrt{n}} diverges:

By DCT, ∑1n+5\sum \frac{1}{\sqrt{n} + 5} diverges.

When DCT is Hard to Use

Sometimes it's hard to prove the inequality!

Example: Is nn2+1\frac{n}{n^2 + 1} less than or greater than 1n\frac{1}{n}?

nn2+1 vs 1n\frac{n}{n^2 + 1} \text{ vs } \frac{1}{n}

Cross-multiply: n2n^2 vs n2+1n^2 + 1

So nn2+1<1n\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 an>0a_n > 0 and bn>0b_n > 0 for all nn.

If: lim⁡n→∞anbn=L\lim_{n \to \infty} \frac{a_n}{b_n} = L

where LL is finite and L>0L > 0, then:

∑an\sum a_n and ∑bn\sum b_n both converge or both diverge!

💡 Key Idea: If anbn→L>0\frac{a_n}{b_n} \to L > 0, then ana_n and bnb_n behave the same way!

Special Cases of LCT

  • If L=0L = 0 and ∑bn\sum b_n converges → ∑an\sum a_n converges
  • If L=∞L = \infty and ∑bn\sum b_n diverges → ∑an\sum a_n diverges
  • If L=0L = 0 and ∑bn\sum b_n diverges → inconclusive
  • If L=∞L = \infty and ∑bn\sum b_n converges → inconclusive

Best to aim for 0<L<∞0 < L < \infty!

How to Use Limit Comparison Test

Step 1: Choose a comparison series ∑bn\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→∞anbn\lim_{n \to \infty} \frac{a_n}{b_n}

Step 3: If 0<L<∞0 < L < \infty, both series have same convergence behavior

Example 3: Use LCT

Determine if ∑n=1∞3n2+5n4−2n+1\sum_{n=1}^{\infty} \frac{3n^2 + 5}{n^4 - 2n + 1} converges.

Step 1: Find dominant terms

Numerator: 3n23n^2 dominates Denominator: n4n^4 dominates

So an≈3n2n4=3n2a_n \approx \frac{3n^2}{n^4} = \frac{3}{n^2}

Compare to bn=1n2b_n = \frac{1}{n^2} (p-series with p=2>1p = 2 > 1, converges!)

Step 2: Compute limit

lim⁡n→∞anbn=lim⁡n→∞3n2+5n4−2n+11n2\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→∞(3n2+5)n2n4−2n+1= \lim_{n \to \infty} \frac{(3n^2 + 5)n^2}{n^4 - 2n + 1}

=lim⁡n→∞3n4+5n2n4−2n+1= \lim_{n \to \infty} \frac{3n^4 + 5n^2}{n^4 - 2n + 1}

Divide by n4n^4:

=lim⁡n→∞3+5n21−2n3+1n4=31=3= \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=3L = 3 (finite and positive), and ∑1n2\sum \frac{1}{n^2} converges.

By LCT, ∑3n2+5n4−2n+1\sum \frac{3n^2 + 5}{n^4 - 2n + 1} converges.

Example 4: Use LCT

Determine if ∑n=1∞n+1n2\sum_{n=1}^{\infty} \frac{\sqrt{n} + 1}{n^2} converges.

Step 1: Dominant terms

Numerator: n=n1/2\sqrt{n} = n^{1/2} Denominator: n2n^2

So an≈n1/2n2=1n3/2a_n \approx \frac{n^{1/2}}{n^2} = \frac{1}{n^{3/2}}

Compare to bn=1n3/2b_n = \frac{1}{n^{3/2}} (p-series with p=32>1p = \frac{3}{2} > 1, converges!)

Step 2: Compute limit

lim⁡n→∞n+1n21n3/2=lim⁡n→∞(n+1)n3/2n2\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→∞n1/2⋅n3/2+n3/2n2= \lim_{n \to \infty} \frac{n^{1/2} \cdot n^{3/2} + n^{3/2}}{n^2}

=lim⁡n→∞n2+n3/2n2= \lim_{n \to \infty} \frac{n^2 + n^{3/2}}{n^2}

Divide by n2n^2:

=lim⁡n→∞(1+1n1/2)=1+0=1= \lim_{n \to \infty} \left(1 + \frac{1}{n^{1/2}}\right) = 1 + 0 = 1

Step 3: Apply LCT

L=1>0L = 1 > 0 and ∑1n3/2\sum \frac{1}{n^{3/2}} converges.

By LCT, the series converges.

Example 5: LCT for Divergence

Determine if ∑n=1∞2n+sin⁡n3n−1\sum_{n=1}^{\infty} \frac{2n + \sin n}{3n - 1} converges.

Step 1: Dominant terms

Numerator: 2n2n dominates (sin n is bounded) Denominator: 3n3n dominates

So an≈2n3n=23a_n \approx \frac{2n}{3n} = \frac{2}{3}

This is like a constant series!

Compare to bn=1b_n = 1 (or any constant)

Actually better: an≈2n3n=23a_n \approx \frac{2n}{3n} = \frac{2}{3}, compare to bn=1n0=1b_n = \frac{1}{n^0} = 1

Wait, we need a series. Since ana_n stays around 23\frac{2}{3} (doesn't go to 0), use nth term test!

lim⁡n→∞2n+sin⁡n3n−1=lim⁡n→∞2+sin⁡nn3−1n=23≠0\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
  • nanb=1nb−a\frac{n^a}{n^b} = \frac{1}{n^{b-a}} (compare to p-series with p=b−ap = b - a)

For exponentials:

  • 2n,en2^n, e^n grow much faster than polynomials
  • Compare to geometric series

For square roots and powers:

  • n=n1/2\sqrt{n} = n^{1/2}
  • n3=n1/3\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: "an<bna_n < b_n and ∑bn\sum b_n diverges, so ∑an\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⁡anbn=0\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 an,bn≥0a_n, b_n \geq 0!

For alternating or negative terms, use different tests.

Mistake 4: Ignoring Smaller Terms

WRONG: "n2+5n+1n3\frac{n^2 + 5n + 1}{n^3} is like n2n3\frac{n^2}{n^3}... wait, what about the 5n5n?"

RIGHT: For large nn, lower powers don't matter! n2n^2 dominates 5n5n and 11.

📝 Practice Strategy

  1. Always try nth term test first: If lim⁡an≠0\lim a_n \neq 0, diverges immediately!
  2. Find dominant terms: Highest powers in numerator and denominator
  3. Choose comparison: Usually p-series ∑1np\sum \frac{1}{n^p} or geometric ∑rn\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: ∑1np\sum \frac{1}{n^p} converges if p>1p > 1, diverges if p≤1p \leq 1

📚 Practice Problems

1Problem 1medium

❓ Question:

Use the Limit Comparison Test to determine if ∑n=1∞n2−5n4+3n2+1\sum_{n=1}^{\infty} \frac{n^2 - 5}{n^4 + 3n^2 + 1} converges.

💡 Show Solution

Step 1: Find dominant terms

Numerator: n2n^2 dominates Denominator: n4n^4 dominates

So: an≈n2n4=1n2a_n \approx \frac{n^2}{n^4} = \frac{1}{n^2}

Compare to bn=1n2b_n = \frac{1}{n^2} (p-series with p=2>1p = 2 > 1, converges)

Step 2: Compute limit

lim⁡n→∞anbn=lim⁡n→∞n2−5n4+3n2+11n2\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{n^2-5}{n^4+3n^2+1}}{\frac{1}{n^2}}

=lim⁡n→∞(n2−5)⋅n2n4+3n2+1= \lim_{n \to \infty} \frac{(n^2-5) \cdot n^2}{n^4+3n^2+1}

=lim⁡n→∞n4−5n2n4+3n2+1= \lim_{n \to \infty} \frac{n^4 - 5n^2}{n^4 + 3n^2 + 1}

Divide by n4n^4:

=lim⁡n→∞1−5n21+3n2+1n4=1−01+0+0=1= \lim_{n \to \infty} \frac{1 - \frac{5}{n^2}}{1 + \frac{3}{n^2} + \frac{1}{n^4}} = \frac{1-0}{1+0+0} = 1

Step 3: Apply LCT

Since L=1>0L = 1 > 0 and ∑1n2\sum \frac{1}{n^2} converges:

By LCT, the series converges.

2Problem 2medium

❓ Question:

Use the Direct Comparison Test to determine whether the series converges or diverges:

∑n=1∞12n+n\sum_{n=1}^{\infty} \frac{1}{2^n + n}

💡 Show Solution

Solution:

Step 1: Find a comparison series.

For large nn, 2n2^n dominates nn, so 2n+n≈2n2^n + n \approx 2^n.

Compare with ∑12n\sum \frac{1}{2^n}, which is a geometric series with r=12<1r = \frac{1}{2} < 1 (converges).

Step 2: Establish the inequality.

For all n≥1n \geq 1: 2n+n>2n2^n + n > 2^n

Therefore: 12n+n<12n\frac{1}{2^n + n} < \frac{1}{2^n}

Step 3: Apply Direct Comparison Test.

We have:

  • 0<12n+n<12n0 < \frac{1}{2^n + n} < \frac{1}{2^n} for all n≥1n \geq 1
  • ∑12n\sum \frac{1}{2^n} converges (geometric series)

By Direct Comparison Test, ∑12n+n\sum \frac{1}{2^n + n} converges.

3Problem 3easy

❓ Question:

Use the Direct Comparison Test to show that ∑n=1∞2+cos⁡nn3\sum_{n=1}^{\infty} \frac{2 + \cos n}{n^3} converges.

💡 Show Solution

Step 1: Find bounds for numerator

Since −1≤cos⁡n≤1-1 \leq \cos n \leq 1:

1≤2+cos⁡n≤31 \leq 2 + \cos n \leq 3

Step 2: Create inequality

0<2+cos⁡nn3≤3n30 < \frac{2 + \cos n}{n^3} \leq \frac{3}{n^3}

Step 3: Check comparison series

∑n=1∞3n3=3∑n=1∞1n3\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>1p = 3 > 1, so it converges.

Step 4: Apply DCT

Since 0<2+cos⁡nn3≤3n30 < \frac{2+\cos n}{n^3} \leq \frac{3}{n^3} and ∑3n3\sum \frac{3}{n^3} converges:

By Direct Comparison Test, the series converges.

4Problem 4hard

❓ Question:

Use the Limit Comparison Test to determine convergence:

∑n=1∞3n2+5n3+n2+1\sum_{n=1}^{\infty} \frac{3n^2 + 5}{n^3 + n^2 + 1}

💡 Show Solution

Solution:

Step 1: Choose a comparison series.

For large nn, the series behaves like 3n2n3=3n\frac{3n^2}{n^3} = \frac{3}{n}.

Compare with ∑1n\sum \frac{1}{n}, the harmonic series (which diverges).

Step 2: Apply Limit Comparison Test.

Let an=3n2+5n3+n2+1a_n = \frac{3n^2 + 5}{n^3 + n^2 + 1} and bn=1nb_n = \frac{1}{n}

L=lim⁡n→∞anbn=lim⁡n→∞3n2+5n3+n2+11nL = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{3n^2 + 5}{n^3 + n^2 + 1}}{\frac{1}{n}}

=lim⁡n→∞n(3n2+5)n3+n2+1= \lim_{n \to \infty} \frac{n(3n^2 + 5)}{n^3 + n^2 + 1}

=lim⁡n→∞3n3+5nn3+n2+1= \lim_{n \to \infty} \frac{3n^3 + 5n}{n^3 + n^2 + 1}

Divide by n3n^3:

=lim⁡n→∞3+5n21+1n+1n3=31=3= \lim_{n \to \infty} \frac{3 + \frac{5}{n^2}}{1 + \frac{1}{n} + \frac{1}{n^3}} = \frac{3}{1} = 3

Step 3: Conclusion.

Since L=3L = 3 (finite and positive) and ∑1n\sum \frac{1}{n} diverges, by the Limit Comparison Test, the original series diverges.

5Problem 5hard

❓ Question:

Determine if ∑n=2∞1nln⁡n−1\sum_{n=2}^{\infty} \frac{1}{n \ln n - 1} converges or diverges.

💡 Show Solution

Step 1: Analyze for large n

For large nn: nln⁡n−1≈nln⁡nn \ln n - 1 \approx n \ln n

So: 1nln⁡n−1≈1nln⁡n\frac{1}{n \ln n - 1} \approx \frac{1}{n \ln n}

Compare to bn=1nln⁡nb_n = \frac{1}{n \ln n}

Step 2: What about ∑1nln⁡n\sum \frac{1}{n \ln n}?

From Integral Test (previous topic), we know:

∑n=2∞1nln⁡n diverges\sum_{n=2}^{\infty} \frac{1}{n \ln n} \text{ diverges}

(The integral ∫1xln⁡xdx=ln⁡∣ln⁡x∣→∞\int \frac{1}{x \ln x}dx = \ln|\ln x| \to \infty)

Step 3: Use Limit Comparison Test

lim⁡n→∞1nln⁡n−11nln⁡n=lim⁡n→∞nln⁡nnln⁡n−1\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 nln⁡nn \ln n:

=lim⁡n→∞11−1nln⁡n=11−0=1= \lim_{n \to \infty} \frac{1}{1 - \frac{1}{n\ln n}} = \frac{1}{1-0} = 1

Step 4: Apply LCT

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

By LCT, the series diverges.

🎴

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