Integral Test 📘

students, in this lesson you will learn one of the most useful tools for deciding whether an infinite series converges or diverges: the Integral Test. This test connects series from Calculus 2 to improper integrals, so it is a powerful bridge between two big ideas. By the end, you should be able to explain when the test works, apply it correctly, and understand how it fits into the broader group of Convergence Tests I.

What the Integral Test Is

A series is an infinite sum, such as $\sum_{n=1}^{\infty} a_n$. In many Calculus 2 problems, we want to know whether this sum approaches a finite value or grows without bound. The Integral Test gives a way to answer that question by comparing the series to an integral.

The test works for series whose terms come from a function $f(x)$ that is:

positive, meaning $f(x) > 0$,
continuous, and
decreasing for all $x \geq N$ for some starting value $N$.

If $a_n = f(n)$, then the Integral Test says that the series $\sum_{n=N}^{\infty} a_n$ and the improper integral $\int_N^{\infty} f(x)\,dx$ either both converge or both diverge.

That means the infinite sum and the infinite-area problem behave the same way. This is a big deal because sometimes the integral is much easier to evaluate or estimate than the series itself. 🌟

A helpful way to think about it is this: if the graph of $f(x)$ is like a set of shrinking positive bars, then the area under the curve and the sum of the bars are closely related.

Why the Integral Test Works

The key idea is geometric. Suppose $f(x)$ is positive and decreasing. Then the rectangles with heights $f(1), f(2), f(3), \dots$ can be compared to the area under the curve $y=f(x)$.

Because the function is decreasing, each rectangle can sit above or below the curve in a controlled way. This creates inequalities that link partial sums of the series to partial integrals of the function. As a result, the series and the improper integral rise or fall together.

More specifically, if $f(x)$ is positive and decreasing, then for integers $n \geq N$:

$\int$_N^{n+1} f(x)\,dx $\leq$ $\sum_{k=N}$^{n} f(k) $\leq$ f(N) + $\int$_N^{n} f(x)\,dx

These comparisons show that the series and integral are tied to the same kind of long-term behavior. If the area under the curve is finite, then the sum of the terms is finite too. If the area is infinite, the series also diverges.

How to Use the Integral Test

To apply the Integral Test, follow these steps carefully:

Identify the function $f(x)$ such that $a_n = f(n)$.
Check the conditions: $f(x)$ must be positive, continuous, and decreasing for large $x$.
Set up the improper integral $\int_N^{\infty} f(x)\,dx$.
Evaluate the integral or determine whether it converges.
Draw the conclusion:

If the integral converges, then the series $\sum_{n=N}^{\infty} a_n$ converges.
If the integral diverges, then the series diverges.

Notice something important: the test does not give the exact value of the series. It only tells whether the series converges or diverges. That makes it a convergence test, not a summation formula.

Also, you do not need the function to be decreasing on every single value of $x$ forever. It is enough if it is decreasing after some point, such as for all $x \geq 3$ or $x \geq 10$. That is often enough for series behavior, because a finite number of early terms does not affect convergence.

Example 1: A Convergent $p$-Series Style Problem ✏️

Consider the series

$\sum_{n=1}^{\infty} \frac{1}{n^2}.$

This is related to the function

$f(x)=\frac{1}{x^2}.$

For $x \geq 1$, the function is positive, continuous, and decreasing, so the Integral Test applies.

Now evaluate the improper integral:

$\int_1^{\infty} \frac{1}{x^2}\,dx.$

Rewrite the integrand as $x^{-2}$:

$\int_1^{\infty} x^{-2}\,dx = \lim_{b\to\infty}\int_1^b x^{-2}\,dx.$

Compute the antiderivative:

$\int x^{-2}\,dx = -x^{-1}.$

So,

$\lim_{b\to\infty}\left[-\frac{1}{x}\right]_1^b$

$= \lim_{b\to\infty}\left(-\frac{1}{b}+1\right)=1.$

Since the integral converges, the series

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

also converges.

This is an example of a $p$-series, which has the form

$\sum_{n=1}^{\infty} \frac{1}{n^p}.$

The Integral Test is one of the ways to prove the standard rule for $p$-series:

if $p>1$, the series converges,
if $p\leq 1$, the series diverges.

Example 2: A Divergent Series

Now consider

$\sum_{n=2}^{\infty} \frac{1}{n}.$

This is the harmonic series, and it is connected to

$ f(x)=\frac{1}{x}.$

For $x \geq 2$, the function is positive, continuous, and decreasing. So we test the integral:

$\int_2^{\infty} \frac{1}{x}\,dx.$

Compute it as an improper integral:

$\int_2^{\infty} \frac{1}{x}\,dx = \lim_{b\to\infty}\int_2^b \frac{1}{x}\,dx.$

Since

$\int \frac{1}{x}\,dx = \ln|x|,$

we get

$\lim_{b\to\infty}\left[\ln x\right]_2^b = \lim_{b\to\infty}(\ln b-\ln 2).$

Because $\ln b \to \infty$ as $b\to\infty$, the integral diverges. Therefore, the harmonic series also diverges.

This is a famous result because the terms $\frac{1}{n}$ get smaller and smaller, but not fast enough to make the infinite sum finite. 📈

Important Details and Common Mistakes

One common mistake is trying to use the Integral Test on a series without checking the conditions. The function must be positive, continuous, and decreasing after some point. If one of those is missing, the test may not apply.

Another mistake is confusing the Integral Test with the Divergence Test. The Divergence Test checks whether $\lim_{n\to\infty} a_n = 0$. If that limit is not $0$, the series diverges immediately. But if the limit is $0$, the series may still converge or diverge. The Integral Test goes further by comparing to an improper integral.

A third mistake is thinking the Integral Test gives the value of the series. It does not. It only tells you whether the series converges or diverges.

Also remember that the lower limit of the sum does not matter much. If you change a series by adding or removing finitely many terms, the convergence behavior stays the same. So starting at $n=1$, $n=2$, or $n=5$ often does not change the answer.

How the Integral Test Fits in Convergence Tests I

In Convergence Tests I, you usually see several first tools for series:

the Divergence Test,
the Integral Test,
and the special class of $p$-series.

These tests work well because they help you make fast decisions about many common series.

The Divergence Test is usually the first check. If

$\lim_{n\to\infty} a_n \neq 0,$

the series diverges right away.

If the terms do go to $0$, then the Integral Test can be useful when the terms come from a nice function $f(x)$. This is especially common for rational functions, logarithmic expressions, and expressions involving powers of $x$.

The Integral Test also leads directly to the behavior of $p$-series. Since $\int_1^{\infty} \frac{1}{x^p}\,dx$ converges exactly when $p>1$, the same is true for the series $\sum_{n=1}^{\infty} \frac{1}{n^p}$. That makes $p$-series a major benchmark in Calculus 2.

So when you face a new series, the big question is: can it be compared to an integral of a positive decreasing function? If yes, the Integral Test may give the answer. ✅

Conclusion

students, the Integral Test is a powerful way to decide whether certain infinite series converge or diverge by comparing them to improper integrals. It works only when the function $f(x)$ is positive, continuous, and decreasing for all sufficiently large $x$. When those conditions are met, the series $\sum a_n$ and the integral $\int f(x)\,dx$ share the same convergence behavior.

This test is important because it connects infinite sums to area under a curve, helps prove the rule for $p$-series, and fits neatly into the first group of convergence tools in Calculus 2. If you can recognize the right function, check the conditions, and evaluate the integral, you will be ready to use one of the classic methods in series analysis.

Study Notes

The Integral Test applies to $\sum_{n=N}^{\infty} a_n$ when $a_n = f(n)$ for a function $f(x)$ that is positive, continuous, and decreasing for $x \geq N$.
If $\int_N^{\infty} f(x)\,dx$ converges, then $\sum_{n=N}^{\infty} a_n$ converges.
If $\int_N^{\infty} f(x)\,dx$ diverges, then $\sum_{n=N}^{\infty} a_n$ diverges.
The Integral Test does not find the exact sum of a series; it only determines convergence or divergence.
The test is especially useful for $p$-series $\sum_{n=1}^{\infty} \frac{1}{n^p}$.
For $p$-series, convergence happens when $p>1$ and divergence happens when $p\leq 1$.
The Divergence Test checks whether $\lim_{n\to\infty} a_n=0$, but that alone does not prove convergence.
A finite number of initial terms does not affect whether a series converges or diverges.
The Integral Test is a key part of Convergence Tests I in Calculus 2.