5.3 The Divergence and Integral Tests

The divergence test and integral test are two of the most frequently used tools for analyzing infinite series. The divergence test gives you a quick way to rule out convergence, while the integral test connects series to improper integrals you already know how to evaluate. Together, they cover a wide range of series you'll encounter in this course.

The Divergence Test

Divergence test for series

The divergence test is the first test you should try on any series. It's fast, and it can immediately tell you a series diverges.

The logic is straightforward: if the terms of a series don't shrink to zero, there's no way the partial sums can settle down to a finite value. Formally:

• If $\lim_{n \to \infty} a_n \neq 0$, or if the limit does not exist, then $\sum_{n=1}^{\infty} a_n$ diverges.
• If $\lim_{n \to \infty} a_n = 0$, the test is inconclusive. The series might converge or might diverge.

That second point trips people up constantly. Having $a_n \to 0$ is necessary for convergence but not sufficient. The classic example is the harmonic series $\sum_{n=1}^{\infty} \frac{1}{n}$: the terms go to zero, yet the series diverges. Meanwhile $\sum_{n=1}^{\infty} \frac{1}{n^2}$ also has terms going to zero, but this one converges. The divergence test can't distinguish between these two cases.

When to use it: Always check this test first. If $\lim_{n \to \infty} a_n \neq 0$, you're done. If the limit equals zero, move on to another test.

The Integral Test

Integral test for convergence

The integral test works by linking a series to an improper integral. Since you already have techniques for evaluating integrals, this lets you leverage that knowledge.

Requirements: To apply the integral test to $\sum_{n=1}^{\infty} a_n$, you need a function $f(x)$ that is:

1. Continuous on $[1, \infty)$
2. Positive on $[1, \infty)$
3. Decreasing on $[1, \infty)$
4. Satisfies $f(n) = a_n$ for all $n \geq 1$

All three conditions must hold (though they only need to hold eventually, meaning for all $x$ beyond some starting point $N$).

The conclusion:

• If $\int_1^{\infty} f(x)\, dx$ converges, then $\sum_{n=1}^{\infty} a_n$ converges.
• If $\int_1^{\infty} f(x)\, dx$ diverges, then $\sum_{n=1}^{\infty} a_n$ diverges.

Note that the integral and the series don't converge to the same value. The test only tells you they share the same convergence behavior.

Example: Does $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converge?

1. Let $f(x) = \frac{1}{x^2}$. This is continuous, positive, and decreasing on $[1, \infty)$.
2. Evaluate $\int_1^{\infty} \frac{1}{x^2}\, dx = \lim_{b \to \infty} \left[-\frac{1}{x}\right]_1^b = \lim_{b \to \infty}\left(-\frac{1}{b} + 1\right) = 1$.
3. The integral converges, so the series converges.

Example: Does $\sum_{n=2}^{\infty} \frac{1}{n \ln n}$ converge?

1. Let $f(x) = \frac{1}{x \ln x}$. This is continuous, positive, and decreasing on $[2, \infty)$.
2. Evaluate using the substitution $u = \ln x$: $\int_2^{\infty} \frac{1}{x \ln x}\, dx = \lim_{b \to \infty} [\ln(\ln x)]_2^b = \infty$.
3. The integral diverges, so the series diverges.

The integral test is especially useful for series involving logarithms, exponentials, or powers where other tests are harder to apply.

Remainder terms for series estimation

When a series converges, you often approximate its value using a partial sum $S_n = \sum_{k=1}^{n} a_k$. The remainder $R_n = S - S_n$ measures how far off that approximation is. The integral test gives you a way to bound this error.

For a series that satisfies the integral test conditions:

$\int_{n+1}^{\infty} f(x)\, dx \leq R_n \leq \int_n^{\infty} f(x)\, dx$

This double inequality is powerful for two reasons:

• Estimating accuracy: If you've computed $S_{10}$ for $\sum \frac{1}{n^2}$, you can bound the error by evaluating $\int_{10}^{\infty} \frac{1}{x^2}\, dx = \frac{1}{10} = 0.1$, so $R_{10} \leq 0.1$.
• Choosing how many terms to sum: If you need the error below $0.01$, set $\int_n^{\infty} f(x)\, dx < 0.01$ and solve for $n$. For $\sum \frac{1}{n^2}$, that gives $\frac{1}{n} < 0.01$, so $n > 100$. You'd need at least 100 terms.

Additional Convergence Tests

Comparison and Limit Comparison Tests

These tests work by comparing a series you don't know to one you do.

Direct Comparison Test: Suppose $0 \leq a_n \leq b_n$ for all $n \geq N$:

• If $\sum b_n$ converges, then $\sum a_n$ converges (a smaller series beneath a convergent one must also converge).
• If $\sum a_n$ diverges, then $\sum b_n$ diverges (a larger series above a divergent one must also diverge).

The tricky part is finding the right series to compare against and getting the inequality in the right direction. Common comparison targets are geometric series and $p$-series.

Limit Comparison Test: If $a_n > 0$ and $b_n > 0$, and

$\lim_{n \to \infty} \frac{a_n}{b_n} = L$

where $0 < L < \infty$, then $\sum a_n$ and $\sum b_n$ either both converge or both diverge. This is often easier than direct comparison because you don't need to establish an inequality; you just need the terms to behave similarly for large $n$.

Telescoping Series

A telescoping series is one where most terms cancel when you write out the partial sums, leaving only a few terms from the beginning and end.

Example: $\sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right)$

Write out the partial sum:

$S_n = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 - \frac{1}{n+1}$

As $n \to \infty$, $S_n \to 1$, so the series converges to 1.

The key step with telescoping series is to use partial fractions (or a similar decomposition) to rewrite the general term as a difference, then verify the cancellation by expanding the partial sum.