Why Match a Series to an Integral?
A series adds values at integer inputs, while an integral measures accumulated area across an interval. When a function is positive and decreases smoothly, the heights of its graph at integers and the areas under its graph are closely related. The Integral Test uses that relationship to decide whether a series converges.
The method has important requirements. It is not enough for the terms to look like values of a function: that function must be positive, continuous, and decreasing on the interval being tested. Checking these conditions first is part of applying the test correctly.
The Integral Test
Suppose a series has terms \(a_n=f(n)\) from some integer \(N\) onward. The Integral Test compares that tail of the series with the improper integral of \(f\) over the matching interval.
Here, \(\int_N^\infty f(x)\,dx\) is an improper integral: by definition, it converges if the finite limit \(\lim_{b\to\infty}\int_N^b f(x)\,dx\) exists. The theorem concerns whether the series converges, not whether its sum equals the integral.
Why the Test Works
Because \(f\) is decreasing, its value at the left endpoint of an interval is at least as large as its values farther to the right. Its value at the right endpoint is at most as large as the values earlier in the interval. These facts give two area comparisons.
The first inequality holds because \(f(x)\leq f(n)\) on \([n,n+1]\), an interval of width \(1\). The second holds because \(f(x)\geq f(n)\) on \([n-1,n]\), also of width \(1\). Adding these comparisons gives, for every integer \(m\geq N\),
The right-hand inequality separates out \(f(N)\); each remaining term \(f(n)\), for \(N+1\leq n\leq m\), is bounded by the area over \([n-1,n]\). Now the equivalence follows. If the improper integral converges, its values \(\int_N^m f(x)\,dx\) are bounded above, so the right-hand inequality bounds all the series’ partial sums. These partial sums are increasing because every term is positive; an increasing, bounded sequence of partial sums converges. Conversely, if the series converges, its partial sums are bounded, so the left-hand inequality bounds \(\int_N^{m+1}f(x)\,dx\) for every \(m\). Since these integrals increase with \(m\), they have a finite limit, and the improper integral converges. The same inequalities show that if the integral grows without bound, the partial sums do too.
A Reliable Application Process
For a series that may fit the test, use this order of checks:
Find a function \(f\) such that \(a_n=f(n)\) for the integer indices in the series tail.
On the relevant interval, establish that \(f\) is positive, continuous, and decreasing.
Write it as a limit of integrals over finite intervals and determine whether that limit is finite.
By the Integral Test, the series and integral either both converge or both diverge.
The hypotheses need to hold only from some point onward. If they hold for \(x\geq N\), the test classifies the tail beginning at \(n=N\). Adding or removing finitely many terms does not change convergence: it changes partial sums by only a finite amount.
Worked Examples
Worked Example: A Convergent Series with an Arctangent Integral
Determine whether \(\sum_{n=1}^{\infty}\frac{1}{n^2+4}\) converges.
Match the terms with \(f(x)=\frac{1}{x^2+4}\) for \(x\geq1\). First, \(x^2+4>0\), so \(f(x)>0\). The denominator is never zero, so \(f\) is continuous. Its derivative is
For \(x\geq1\), the numerator is negative and the denominator is positive, so \(f'(x)<0\). Thus \(f\) is decreasing, and all three hypotheses hold.
An antiderivative is \(\frac12\arctan(x/2)\): differentiating gives \(\frac12\cdot\frac{1/2}{1+(x/2)^2}=\frac{1}{x^2+4}\). Therefore
This is finite, since both arctangent values are finite. The Integral Test therefore proves that \(\sum_{n=1}^{\infty}\frac{1}{n^2+4}\) converges. The antiderivative check confirms the integral calculation, and the verified hypotheses justify the conclusion.
Worked Example: A Divergent Series with a Logarithmic Integral
Determine whether \(\sum_{n=1}^{\infty}\frac{n}{n^2+1}\) converges.
Use \(f(x)=\frac{x}{x^2+1}\) for \(x\geq1\). The denominator is positive, so \(f(x)>0\), and it has no zeros, so \(f\) is continuous. By the quotient rule,
For \(x\geq1\), \(1-x^2\leq0\), so \(f'(x)\leq0\) and \(f\) is decreasing. In particular, it is allowed that \(f'(1)=0\). The three hypotheses hold.
An antiderivative is \(\frac12\ln(x^2+1)\), because its derivative is \(\frac12\cdot\frac{2x}{x^2+1}=\frac{x}{x^2+1}\). Hence
As \(b\to\infty\), \(\frac{b^2+1}{2}\to\infty\), so the logarithm tends to infinity. The improper integral diverges. By the Integral Test, the series diverges. The derivative and antiderivative checks confirm the steps used to reach this conclusion.
Worked Example: An Exponential Series
Determine whether \(\sum_{n=1}^{\infty}e^{-n}\) converges.
Let \(f(x)=e^{-x}\) for \(x\geq1\). The exponential is positive and continuous, and
Thus \(f\) is decreasing, so the test’s three hypotheses are satisfied. An antiderivative is \(-e^{-x}\), whose derivative is \(e^{-x}\). It follows that
The integral is finite, so the series converges by the Integral Test. Substituting the limits into the antiderivative gives the stated value, and the derivative check verifies that it is an antiderivative. This conclusion is about convergence; the Integral Test does not claim that the series sum is \(e^{-1}\).
Common Pitfall: The Integral Is Not the Series Sum
A frequent error is to evaluate the improper integral and declare that its value equals the sum of the series. The Integral Test does not establish that equality. It establishes only that the series and integral have the same convergence behavior.
For example, \(f(x)=\frac{1}{x^2+4}\) is strictly decreasing for \(x\geq1\). On every interval \([n,n+1]\), \(f(x)<f(n)\) for \(x>n\), so
Summing over intervals shows that \(\int_1^\infty f(x)\,dx\) is less than \(\sum_{n=1}^\infty f(n)\), not equal to it. The integral’s finite value proves convergence here, but it is not the series sum. Always state the conclusion the test supports: convergence or divergence.
Another mistake is to skip checking the hypotheses. A formula that produces positive terms at integer inputs does not automatically define a positive, continuous, decreasing function over the whole interval. If the hypotheses fail, this test cannot be used as written; that failure alone does not determine whether the series converges.
Why the Integral Test Matters
The test turns a question about infinitely many discrete terms into a question about the total area under a curve. The proof explains why this works: decreasingness traps each term between areas of neighboring unit intervals. Positivity ensures that both the partial sums and the accumulated areas increase, so their boundedness or unboundedness determines convergence or divergence.
When applying the test, make the matching explicit, verify positivity, continuity, and decrease on the interval, and then evaluate the improper integral as a limit. Keeping those steps separate makes the conclusion clear and prevents an invalid use of the test.
Check Your Understanding
For each question, identify the hypothesis or comparison that justifies your answer.
- State the three conditions on \(f\) required to apply the Integral Test to \(a_n=f(n)\).
- Why does the theorem allow the hypotheses to hold only for \(x\geq N\), rather than from \(x=1\)?
- For a positive decreasing function, which is larger: \(f(n)\) or \(\int_n^{n+1}f(x)\,dx\)? Explain using the interval’s width.
- If the matching improper integral diverges and the Integral Test hypotheses hold, what follows about the series?
- Does the Integral Test say that a convergent series has the same sum as its matching integral? Explain.
