The Main Idea 

The divergence test is the simplest and most fundamental test for series convergence. It’s based on a logical necessity: if you’re adding infinitely many terms and want a finite sum, those terms must be getting smaller and smaller—specifically, they must approach zero.

If [latex]\sum_{n=1}^{\infty} a_n[/latex] converges to [latex]S[/latex], then consecutive partial sums [latex]S_k[/latex] and [latex]S_{k-1}[/latex] both approach [latex]S[/latex]. Since [latex]a_k = S_k - S_{k-1}[/latex], we must have [latex]a_k \to S - S = 0[/latex].

The test: If [latex]\lim_{n \to \infty} a_n \neq 0[/latex] (or the limit doesn’t exist), then [latex]\sum_{n=1}^{\infty} a_n[/latex] diverges.

The divergence test can only prove divergence, never convergence. When [latex]\lim_{n \to \infty} a_n = 0[/latex], the test is inconclusive—the series might converge or diverge.

Always apply the divergence test first! It’s quick and can immediately rule out many series. Look for terms that don’t approach zero, like [latex]\frac{n}{3n-1} \to \frac{1}{3}[/latex] or [latex]e^{\frac{1}{n^2}} \to e^0 = 1[/latex].

The Main Idea 

The integral test provides a powerful way to determine series convergence by comparing an infinite sum to an improper integral. Since both represent “adding up” infinitely many values, their convergence behaviors are linked.

For a series [latex]\sum_{n=1}^{\infty} a_n[/latex] with positive terms, find a continuous, decreasing function [latex]f(x)[/latex] where [latex]f(n) = a_n[/latex]. Then [latex]\sum_{n=1}^{\infty} a_n[/latex] and [latex]\int_1^{\infty} f(x)dx[/latex] either both converge or both diverge.

Think of rectangles representing series terms compared to the area under a curve. When the rectangles sit above the curve (like with [latex]\frac{1}{n}[/latex] and [latex]\frac{1}{x}[/latex]), if the integral diverges, the series must also diverge. When rectangles sit below the curve (like with [latex]\frac{1}{n^2}[/latex] and [latex]\frac{1}{x^2}[/latex]), if the integral converges, the series is bounded and must converge.

Key requirements:

• Series must have positive terms
• Function [latex]f(x)[/latex] must be continuous, positive, and decreasing
• [latex]f(n) = a_n[/latex] for all sufficiently large [latex]n[/latex]

The integral test only tells you about convergence/divergence, not the actual sum. The series sum and integral value are generally different (like [latex]\sum(\frac{1}{e})^n = \frac{1}{e-1}[/latex] versus [latex]\int_1^{\infty}(\frac{1}{e})^x dx = \frac{1}{e}[/latex]).

The Main Idea 

The p-series [latex]\sum_{n=1}^{\infty} \frac{1}{n^p}[/latex] represents one of the most fundamental families of series in mathematics. Understanding p-series gives you a benchmark for testing many other series through comparison techniques.

The simple convergence rule: The p-series [latex]\sum_{n=1}^{\infty} \frac{1}{n^p}[/latex] converges if and only if [latex]p > 1[/latex].

Why this cutoff at [latex]p = 1[/latex]:

• When [latex]p \leq 0[/latex]: Terms don’t approach zero, so divergence test kills it immediately
• When [latex]0 < p < 1[/latex]: Terms approach zero too slowly—integral test with [latex]\int_1^{\infty} \frac{1}{x^p}dx = \infty[/latex] shows divergence
• When [latex]p = 1[/latex]: This is the harmonic series, which diverges (borderline case)
• When [latex]p > 1[/latex]: Terms approach zero fast enough—integral test with [latex]\int_1^{\infty} \frac{1}{x^p}dx = \frac{1}{p-1}[/latex] shows convergence

The value [latex]p = 1[/latex] represents the boundary between convergence and divergence. The harmonic series shows that even when terms approach zero, they must do so “fast enough”—faster than [latex]\frac{1}{n}[/latex].

The Main Idea 

When a series converges, you often want to know its actual sum, not just that it converges. The integral test provides a powerful way to estimate how close your partial sums are to the true infinite sum.

The remainder concept: For a convergent series [latex]\sum_{n=1}^{\infty} a_n[/latex], the remainder [latex]R_N = \sum_{n=1}^{\infty} a_n - S_N[/latex] represents the “error” when you approximate the infinite sum with the [latex]N[/latex]th partial sum [latex]S_N[/latex].

The key insight: If your series satisfies the integral test conditions, then: [latex]\int_{N+1}^{\infty} f(x)dx < R_N < \int_N^{\infty} f(x)dx[/latex]

This gives you concrete upper and lower bounds on your approximation error.

Problem-Solving Strategy:

1. Calculate [latex]S_N[/latex] for some reasonable [latex]N[/latex]
2. Evaluate [latex]\int_N^{\infty} f(x)dx[/latex] to get an upper bound on the error
3. Add this bound to [latex]S_N[/latex] to get an overestimate of the series sum
4. If you need the error below a certain threshold, solve [latex]\int_N^{\infty} f(x)dx < \text{desired error}[/latex] for [latex]N[/latex]

The integral test doesn’t just tell you about convergence—it quantifies how quickly the series approaches its limit, letting you control approximation accuracy.