Why the exponent determines convergence
A p-series has terms that become smaller as the positive integer index grows, but that fact alone does not decide whether their sum is finite. The exponent \(p\) determines how quickly the terms shrink. The integral test makes the boundary between convergence and divergence visible by replacing the discrete terms with the function \(x^{-p}\).
The key is to evaluate the area accumulated under this function from \(1\) onward. If that improper integral is finite, the integral test says the p-series converges; if the integral is infinite, the series diverges. The exponent \(p=1\) is a special case because its antiderivative is logarithmic rather than a power function.
Set up the integral test
The integral test applies when a function \(f\) is positive, continuous, and decreasing on \([1,\infty)\), and the series has terms \(f(n)\). Under those conditions, the series \(\sum_{n=1}^{\infty}f(n)\) and the improper integral \(\int_1^\infty f(x)\,dx\) either both converge or both diverge.
For a p-series, take \(f(x)=x^{-p}\), with \(p>0\). This function is positive and continuous for \(x\geq 1\). Its derivative is
because both \(p\) and \(x^{-p-1}\) are positive. Thus \(f\) is decreasing, so the integral test applies. We need to determine whether
has a finite value. The limit in this definition is essential: an integral over an unbounded interval is called convergent only if the limit of the finite-interval integrals exists as a finite number.
Evaluate the improper integral
When \(p\neq 1\), the power rule gives an antiderivative \(x^{1-p}/(1-p)\). Therefore, for finite \(b>1\),
When \(p=1\), the integrand is \(1/x\), whose antiderivative is \(\ln x\). Hence
These expressions have different limits according to the exponent. For \(p>1\), the exponent \(1-p\) is negative, so \(b^{1-p}\) tends to zero as \(b\to\infty\). For \(0<p<1\), the exponent \(1-p\) is positive, so \(b^{1-p}\) grows without bound. At \(p=1\), \(\ln b\) also grows without bound.
Why the threshold is exactly one
The theorem follows by combining the integral evaluation with the conditions of the integral test. The argument treats the three possible ranges separately, so the boundary case is not hidden inside a general power formula.
For every \(p>0\), \(f(x)=x^{-p}\) is positive and continuous on \([1,\infty)\), and its derivative is negative there. Thus \(f\) is decreasing and the integral test applies.
Here \(1-p<0\), so \(b^{1-p}\to 0\). Consequently, the improper integral equals \(\frac{0-1}{1-p}=\frac{1}{p-1}\), which is finite. The integral test implies that the p-series converges.
Here \(1-p>0\), so \(b^{1-p}\to\infty\). The expression \(\frac{b^{1-p}-1}{1-p}\) therefore tends to infinity. The integral test implies that the p-series diverges.
The integral is \(\lim_{b\to\infty}\ln b=\infty\). The integral test therefore implies that the p-series diverges at \(p=1\) as well.
This proves both directions of the classification: every \(p>1\) gives a convergent series, and every positive \(p\leq 1\) gives a divergent series. The case \(p=1\) belongs with the divergence range, not the convergence range.
Worked examples at and around the threshold
Worked Example: The p-series with \(p=2\)
Classify \(\sum_{n=1}^{\infty}1/n^2\) using the integral test. Here \(f(x)=x^{-2}\). It is positive and continuous for \(x\geq1\), and \(f'(x)=-2x^{-3}<0\), so the test applies.
For a finite upper endpoint \(b\), an antiderivative is \(-x^{-1}\). Thus
Taking the limit gives \(\lim_{b\to\infty}(1-1/b)=1\), which is finite. Therefore the series converges. Check the antiderivative by differentiating: \(\frac{d}{dx}(-x^{-1})=x^{-2}\). The integral calculation and the integral test both support the convergence conclusion.
Worked Example: The boundary case \(p=1\)
Classify the harmonic series \(\sum_{n=1}^{\infty}1/n\). Set \(f(x)=1/x\). The function is positive and continuous on \([1,\infty)\), and \(f'(x)=-1/x^2<0\), so the integral test applies.
For finite \(b>1\),
As \(b\to\infty\), \(\ln b\to\infty\). The improper integral diverges, so the harmonic series diverges by the integral test. Check the antiderivative: \(\frac{d}{dx}\ln x=1/x\) for \(x>0\). In particular, the boundary exponent \(p=1\) does not produce convergence.
Worked Example: A p-series with \(0<p<1\)
Classify \(\sum_{n=1}^{\infty}1/\sqrt n\). Since \(1/\sqrt n=1/n^{1/2}\), take \(p=1/2\) and \(f(x)=x^{-1/2}\). It is positive and continuous on \([1,\infty)\), and \(f'(x)=-\tfrac12 x^{-3/2}<0\), so the integral test applies.
An antiderivative is \(2\sqrt{x}\). Therefore,
Since \(\sqrt b\to\infty\), this integral tends to infinity. The series diverges by the integral test. Check the antiderivative: \(\frac{d}{dx}(2\sqrt{x})=2\cdot\frac{1}{2\sqrt{x}}=x^{-1/2}\). This example confirms that having \(p>0\) is not enough for convergence: the exponent must exceed one.
What the integral test does—and does not—tell you
The integral test gives a classification, not usually an exact value for the series. In the first example, the integral \(\int_1^\infty x^{-2}\,dx\) equals \(1\). That does not mean the sum \(\sum_{n=1}^{\infty}1/n^2\) equals \(1\). The test states that the series and integral share convergence or divergence; it does not state that their values are equal.
For \(p>1\), the integral does have the exact value \(1/(p-1)\). This is the value of the improper integral, not automatically the value of the p-series. For \(0<p\leq1\), the integral is infinite, which is enough to classify the series as divergent.
A second mistake is to equate the series sum with the integral. For \(p=2\), the integral is \(1\), but the series includes the first term \(1\) and also has further positive terms, so its sum cannot equal \(1\). The correct conclusion from the integral test is only that the series converges.
Why this threshold is useful
The p-series is a standard reference point for judging how quickly terms decrease. Its threshold gives a clear distinction: powers just above one produce convergence, while powers equal to or below one produce divergence. When examining a new positive-term series, recognizing a p-series can therefore provide an immediate benchmark for its rate of decay.
The method also illustrates an important role of improper integrals. A finite integral measures a finite total accumulation under the continuous function, while an unbounded integral signals unlimited accumulation. The integral test transfers that conclusion to the corresponding discrete sum, provided its positivity, continuity, and decrease conditions hold.
Check Your Understanding
Use the integral-test reasoning in this tutorial to answer each question.
- For which positive values of \(p\) does \(\sum_{n=1}^{\infty}1/n^p\) converge?
- Why does the integral test apply to \(f(x)=x^{-p}\) when \(p>0\)?
- Evaluate \(\int_1^b x^{-p}\,dx\) for \(p\neq1\), and state how its limit depends on whether \(p>1\) or \(0<p<1\).
- Why must \(p=1\) be handled separately from the power-antiderivative formula?
- If \(\int_1^\infty f(x)\,dx\) converges under the conditions of the integral test, what does that establish about \(\sum_{n=1}^{\infty}f(n)\), and what does it not establish about its exact sum?