The Main Idea 

The comparison test leverages series you already understand (like geometric and p-series) to determine convergence of new, similar-looking series. It’s based on a simple principle: if you’re smaller than something that converges, you converge; if you’re bigger than something that diverges, you diverge.

The test has two parts:

• For convergence: If [latex]0 \leq a_n \leq b_n[/latex] and [latex]\sum b_n[/latex] converges, then [latex]\sum a_n[/latex] converges
• For divergence: If [latex]a_n \geq b_n \geq 0[/latex] and [latex]\sum b_n[/latex] diverges, then [latex]\sum a_n[/latex] diverges

Why does this work? Series with positive terms have monotone increasing partial sums. If your series is term-by-term smaller than a convergent series, its partial sums are bounded above and must converge. If your series is term-by-term larger than a divergent series, its partial sums are unbounded and must diverge.

Strategy for choosing comparison series:

• For convergence: Find a larger, known convergent series (often a convergent p-series like [latex]\sum \frac{1}{n^2}[/latex] or geometric series)
• For divergence: Find a smaller, known divergent series (often the harmonic series [latex]\sum \frac{1}{n}[/latex] or a divergent p-series)

Look at the dominant terms. For [latex]\frac{1}{n^3 + 3n + 1}[/latex], the [latex]n^3[/latex] term dominates for large [latex]n[/latex], so compare with [latex]\frac{1}{n^3}[/latex]. For [latex]\frac{1}{\ln n}[/latex], since [latex]\ln n < n[/latex], we have [latex]\frac{1}{\ln n} > \frac{1}{n}[/latex].

Common comparison series:

• p-series: [latex]\sum \frac{1}{n^p}[/latex] (converges if p > 1, diverges if p ≤ 1)
• Geometric: [latex]\sum ar^{n-1}[/latex] (converges if |r| < 1)
• Harmonic: [latex]\sum \frac{1}{n}[/latex] (diverges)

The comparison test only works when you can establish the right inequality. Sometimes the comparison isn’t obvious, and you need other tests.

The Main Idea 

The limit comparison test handles situations where the regular comparison test can’t help—when your series and a known series have similar behavior, but the inequalities go the “wrong” way for a direct comparison.

Instead of comparing terms directly, compare their ratio as [latex]n \to \infty[/latex]. If [latex]\lim_{n \to \infty} \frac{a_n}{b_n} = L[/latex] where [latex]L[/latex] is finite and nonzero, then [latex]\sum a_n[/latex] and [latex]\sum b_n[/latex] have the same convergence behavior.

The three cases:

• Case 1: [latex]\lim_{n \to \infty} \frac{a_n}{b_n} = L \neq 0[/latex] (finite, nonzero) → Both series converge or both diverge
• Case 2: [latex]\lim_{n \to \infty} \frac{a_n}{b_n} = 0[/latex] → If [latex]\sum b_n[/latex] converges, then [latex]\sum a_n[/latex] converges
• Case 3: [latex]\lim_{n \to \infty} \frac{a_n}{b_n} = \infty[/latex] → If [latex]\sum b_n[/latex] diverges, then [latex]\sum a_n[/latex] diverges

When to use it: Perfect for series with similar “dominant behavior”—like [latex]\frac{1}{n^2-1}[/latex] compared to [latex]\frac{1}{n^2}[/latex], where both behave essentially like [latex]\frac{1}{n^2}[/latex] for large [latex]n[/latex].

Look at the dominant terms in numerator and denominator. For [latex]\frac{1}{\sqrt{n}+1}[/latex], the dominant behavior is [latex]\frac{1}{\sqrt{n}}[/latex], so compare with [latex]\sum \frac{1}{\sqrt{n}}[/latex].

When the test fails: Cases 2 and 3 can be inconclusive if you pick the wrong comparison series. If [latex]\frac{a_n}{b_n} \to 0[/latex] but [latex]\sum b_n[/latex] diverges, or if [latex]\frac{a_n}{b_n} \to \infty[/latex] but [latex]\sum b_n[/latex] converges, try a different comparison series.