Integral Test for P-Series and Logarithmic Terms

Why These Series Have Important Endpoints

Many series have terms that shrink as \(n\) grows, yet their sums behave differently. The p-series \(\sum 1/n^p\) and the logarithmic series \(\sum 1/[n(\ln n)^q]\) are useful models: changing an exponent can change whether the total sum is finite. The Integral Test connects these questions about sums to improper integrals that can be evaluated directly.

The endpoint values matter. For a p-series, the dividing exponent is \(p=1\); for a logarithmic series of this form, it is \(q=1\). At either endpoint, substituting into a formula meant for other exponents can give a false conclusion. We will derive the classifications and check the boundary cases explicitly.

The Integral Test and Its Conditions

Theorem: Integral Test. Suppose \(f\) is continuous, positive, and decreasing on \([N,\infty)\), where \(N\) is a positive integer, and suppose \(a_n=f(n)\) for integers \(n\ge N\). Then \(\sum_{n=N}^{\infty}a_n\) converges if and only if the improper integral \(\int_N^\infty f(x)\,dx\) converges. If the integral diverges, the series diverges. Changing or omitting finitely many terms does not affect convergence.

The conditions are part of the theorem, not optional checks. In particular, the function must be positive and decreasing on the interval where the comparison is made. A function need only satisfy these conditions from some sufficiently large \(N\): any earlier terms contribute a finite amount and cannot change whether the tail has a finite sum.

Why the Integral Test Works

If \(f\) is decreasing, then its graph on each interval between consecutive integers lies below the left-endpoint value and above the right-endpoint value. For integers \(n\ge N\), this gives

\int_n^{n+1} f(x)\,dx\le f(n), \qquad f(n)\le \int_{n-1}^{n} f(x)\,dx.

Adding these inequalities across consecutive intervals gives, for \(M\ge N\),

\int_N^{M+1} f(x)\,dx \le \sum_{n=N}^{M}f(n) \le f(N)+\int_N^{M} f(x)\,dx.

These bounds explain both directions of the test. If the integral has a finite limit as its upper endpoint grows, the right-hand bound keeps the partial sums bounded, so the positive-term series converges. If the integral grows without bound, the left-hand bound forces the partial sums to grow without bound, so the series diverges.

Choose a function.
Write the series term as \(a_n=f(n)\) and identify a suitable interval.

Check the hypotheses.
Verify that \(f\) is continuous, positive, and decreasing on the chosen interval.

Evaluate the improper integral.
Find whether its limit is finite or infinite as the upper endpoint tends to infinity.

Transfer the result to the series.
Use the Integral Test to classify convergence or divergence.

The p-Series Classification

A p-series is a series of the form \(\sum_{n=1}^{\infty}1/n^p\), where \(p\) is a real number. For \(p>0\), let \(f(x)=x^{-p}\) on \([1,\infty)\). It is continuous and positive there, and

f'(x)=-p x^{-p-1}<0.

Thus \(f\) is decreasing, so the Integral Test applies. For \(p\ne 1\), an antiderivative is \(x^{1-p}/(1-p)\). For \(p=1\), the antiderivative is \(\ln x\). Evaluating the resulting improper integrals gives the complete classification:

Theorem: p-Series Test. The series \(\sum_{n=1}^{\infty}1/n^p\) converges exactly when \(p>1\), and diverges when \(p\le 1\). For \(p\le 0\), its terms do not approach zero, so divergence follows from the nth-term test rather than the Integral Test.

Indeed, if \(p>1\), then \(1-p<0\), so \(b^{1-p}\) tends to zero and the integral from \(1\) to infinity is finite. If \(0<p<1\), then \(1-p>0\), so the integral grows without bound. At \(p=1\), it is \(\ln b\), which also grows without bound. This identifies the endpoint rather than inferring it from nearby cases.

Worked Example: A Convergent p-Series

Classify \(\sum_{n=1}^{\infty}1/n^{3/2}\). Set \(f(x)=x^{-3/2}\). It is continuous and positive for \(x\ge1\), and

f'(x)=-\frac{3}{2}x^{-5/2}<0.

The function is decreasing, so we evaluate the improper integral:

\int_1^\infty x^{-3/2}\,dx =\lim_{b\to\infty}\left[-2x^{-1/2}\right]_1^b =\lim_{b\to\infty}\left(2-\frac{2}{\sqrt b}\right) =2.

The antiderivative checks because \(\frac{d}{dx}(-2x^{-1/2})=x^{-3/2}\). The integral is finite, so the Integral Test shows that the series converges. This agrees with the p-series condition \(3/2>1\).

Worked Example: The Harmonic Endpoint

Classify \(\sum_{n=1}^{\infty}1/n\). Take \(f(x)=1/x\), which is continuous, positive, and decreasing for \(x\ge1\). Its improper integral is

\int_1^\infty \frac{1}{x}\,dx =\lim_{b\to\infty}[\ln x]_1^b =\lim_{b\to\infty}\ln b =\infty.

The antiderivative checks because \(\frac{d}{dx}\ln x=1/x\). Since the integral diverges, the Integral Test shows that the harmonic series diverges. Here \(p=1\), exactly the boundary value of the p-series classification.

Logarithmic Terms and Their Endpoint

Now consider \(\sum_{n=2}^{\infty}1/[n(\ln n)^q]\), where \(q\) is real. Starting at \(n=2\) ensures that \(\ln n\) is positive, so the real power \((\ln n)^q\) is defined and positive. Use \(f(x)=1/[x(\ln x)^q]\) for \(x>1\). This function is continuous and positive. Its derivative can be written as

f'(x) = -\frac{\ln x+q}{x^2(\ln x)^{q+1}}.

For any fixed real \(q\), choose \(N\) large enough that \(\ln N>\max(0,-q)\). Then \(\ln x+q>0\) for \(x\ge N\), and the denominator in the derivative is positive. Thus \(f'(x)<0\) on that tail, and the Integral Test applies there. Any terms from \(n=2\) up to \(N-1\) are a finite initial segment and do not affect the classification.

To evaluate the tail integral, substitute \(u=\ln x\), so \(du=dx/x\). The integral becomes an integral of \(u^{-q}\). When \(q\ne1\), an antiderivative is \(u^{1-q}/(1-q)\); when \(q=1\), it is \(\ln u\). Therefore the series converges exactly when \(q>1\). At \(q=1\), the integral grows like \(\ln(\ln b)\), which still tends to infinity.

Theorem: Logarithmic Series Test. For real \(q\), the series \(\sum_{n=2}^{\infty}1/[n(\ln n)^q]\) converges if \(q>1\) and diverges if \(q\le1\). The function in the Integral Test need only be decreasing eventually; a finite number of terms before that point does not change the result.

Worked Example: A Convergent Logarithmic Series

Classify \(\sum_{n=2}^{\infty}1/[n(\ln n)^2]\). Here \(q=2\), and

f'(x) = -\frac{\ln x+2}{x^2(\ln x)^3}<0 \qquad (x>1).

The function is continuous, positive, and decreasing on \([2,\infty)\). With \(u=\ln x\) and \(du=dx/x\), its integral is

\int_2^\infty\frac{dx}{x(\ln x)^2} = \int_{\ln 2}^\infty u^{-2}\,du = \lim_{B\to\infty}\left[-\frac{1}{u}\right]_{\ln 2}^{B} = \frac{1}{\ln 2}.

The antiderivative checks: \(\frac{d}{dx}[-1/\ln x]=1/[x(\ln x)^2]\). Since \(1/\ln 2\) is finite, the Integral Test proves that the series converges. The threshold check also gives \(q=2>1\).

Worked Example: The Logarithmic Endpoint

Classify \(\sum_{n=2}^{\infty}1/[n\ln n]\). Set \(q=1\). The function \(f(x)=1/(x\ln x)\) is continuous and positive for \(x\ge2\), and

f'(x) = -\frac{\ln x+1}{x^2(\ln x)^2}<0 \qquad (x\ge2).

The integral is

\int_2^\infty\frac{dx}{x\ln x} = \int_{\ln 2}^\infty\frac{du}{u} = \lim_{B\to\infty}[\ln u]_{\ln 2}^{B} = \lim_{B\to\infty}\bigl(\ln B-\ln(\ln 2)\bigr) = \infty.

Equivalently, an antiderivative in \(x\) is \(\ln(\ln x)\), whose derivative is \(1/(x\ln x)\). The integral diverges, so the series diverges. The endpoint \(q=1\) belongs to the divergent range \(q\le1\).

A Common Endpoint Mistake

A tempting error is to use the antiderivative \(x^{1-p}/(1-p)\) at \(p=1\). Substitution makes the denominator \(1-p\) equal to zero, so the formula is not defined. One might incorrectly conclude that the integral is finite by treating the zero denominator as an algebraic detail. It is not: the antiderivative changes at this exponent.

The correction is to evaluate the endpoint case separately:

\int_1^b x^{-1}\,dx=\ln b, \qquad \lim_{b\to\infty}\ln b=\infty.

The same care is needed for logarithmic series. The formula \(u^{1-q}/(1-q)\) does not apply at \(q=1\); there the antiderivative is \(\ln u\), producing divergence. Always check whether a parameter value makes a denominator in an antiderivative zero, and handle that value directly.

Why the Classifications Are Useful

These two families make clear that a small change in an exponent can alter whether a sum is finite. For p-series, powers greater than one converge, while the harmonic boundary and all smaller powers diverge. When the leading factor is \(1/n\), a power of \(\ln n\) must have exponent strictly greater than one to make the series converge. Neither endpoint can be included in the convergent range.

For the Integral Test, keep the reasoning in order: identify the function, check positivity and eventual decrease, evaluate the improper integral, then transfer its behavior to the series. The test establishes convergence or divergence; the value of the integral is not, in general, the exact value of the series.

Check Your Understanding

For each question, state the relevant test or endpoint reasoning, not just the conclusion.

Does \(\sum_{n=1}^{\infty}1/n^{4/3}\) converge or diverge? Evaluate the corresponding improper integral.
Does \(\sum_{n=1}^{\infty}1/\sqrt n\) converge or diverge? What happens to \(\int_1^b x^{-1/2}\,dx\) as \(b\to\infty\)?
Classify \(\sum_{n=2}^{\infty}1/[n(\ln n)^{5/4}]\), and state the substitution used in its integral.
Why does \(\sum_{n=2}^{\infty}1/[n\ln n]\) diverge even though its terms approach zero?
For a real \(q\), why is it enough that \(1/[x(\ln x)^q]\) be decreasing only for all sufficiently large \(x\)?

Integral Test for P-Series and Logarithmic Terms

What You'll Learn

Why These Series Have Important Endpoints

The Integral Test and Its Conditions

Why the Integral Test Works