Mixed Convergence and Divergence Problems

Classifying Series with More Than One Test

A series with positive terms can often be classified by comparing it with a familiar series or by evaluating an improper integral. The central task is not to use every available test. It is to identify a test whose hypotheses can be checked and whose comparison or integral leads to a definite conclusion.

In the examples below, the terms will be positive. That matters: the comparison rules used here rely on inequalities between nonnegative terms. A useful reference point is the \(p\)-series \(\sum_{n=1}^{\infty}1/n^p\), which converges when \(p>1\) and diverges when \(p\le 1\), for \(p>0\). The harmonic series \(\sum_{n=1}^{\infty}1/n\) is the case \(p=1\).

Definition: A series \(\sum_{n=N}^{\infty}a_n\) with \(a_n\ge 0\) converges if its partial sums have a finite limit. It diverges if those partial sums do not have a finite limit. Changing or removing finitely many terms does not change whether a series converges.

Two Comparison Rules

Direct comparison uses inequalities between terms. The direction of the inequality determines which conclusion is available.

Theorem: Suppose \(a_n\ge 0\) and \(b_n\ge 0\) for all sufficiently large \(n\). If \(a_n\le b_n\) and \(\sum b_n\) converges, then \(\sum a_n\) converges. If \(a_n\ge b_n\) and \(\sum b_n\) diverges, then \(\sum a_n\) diverges. The inequalities need only hold from some index onward.

The first rule bounds the partial sums of \(\sum a_n\) above by partial sums of a convergent positive-term series. The second rule bounds partial sums of \(\sum a_n\) below by partial sums that grow without bound. A comparison in the opposite direction does not establish the corresponding conclusion.

The integral test applies when the terms are values of a suitable function.

Theorem: Suppose \(f\) is positive, continuous, and decreasing on \([N,\infty)\), and \(a_n=f(n)\) for integers \(n\ge N\). Then \(\sum_{n=N}^{\infty}a_n\) and \(\int_N^{\infty}f(x)\,dx\) either both converge or both diverge.

Why the Integral Test Works

Because \(f\) is decreasing, its values on each interval \([n,n+1]\) lie between its values at the endpoints. Thus, for every integer \(n\ge N\),

f(n+1)\le f(x)\le f(n) \qquad (n\le x\le n+1).

Integrating these inequalities over an interval of length one gives

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

Adding these inequalities for \(n=N,\ldots,M\) yields

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

These bounds explain both directions of the test. If the improper integral is finite, the left inequality bounds the series tail \(\sum_{n=N+1}^{M+1}f(n)\) by a finite integral, so the increasing partial sums of the positive-term series are bounded and have a finite limit. If the series converges, the right inequality shows that the integrals \(\int_N^{M+1}f(x)\,dx\) are bounded above by the partial sums, so the increasing integrals have a finite limit. Therefore the integral and series have the same convergence behavior. Positivity and decreasing behavior are essential to these bounds.

Worked Examples: Applying the Tests

Worked Example: A Convergent Series by the Integral Test

Classify \(\sum_{n=1}^{\infty}\frac{n}{n^3+1}\).

Set \(f(x)=\frac{x}{x^3+1}\) for \(x\ge1\). The denominator is positive, so \(f(x)>0\); the function is continuous on \([1,\infty)\). Differentiate using the quotient rule:

f'(x) =\frac{(x^3+1)(1)-x(3x^2)}{(x^3+1)^2} =\frac{1-2x^3}{(x^3+1)^2}.

For \(x\ge1\), \(1-2x^3\le -1<0\), while the denominator is positive. Thus \(f'(x)<0\), so \(f\) is decreasing. All the integral-test hypotheses hold.

For \(x\ge1\), \(x^3+1\ge x^3\), and therefore

0<f(x)=\frac{x}{x^3+1} \le \frac{x}{x^3} =\frac{1}{x^2}.

Consequently, for every \(R>1\),

0\le \int_1^R\frac{x}{x^3+1}\,dx \le \int_1^R\frac{1}{x^2}\,dx =1-\frac{1}{R}<1.

The improper integral is finite, since its partial integrals are increasing and bounded above by \(1\). The integral test now gives

\sum_{n=1}^{\infty}\frac{n}{n^3+1} \quad\text{converges.}

Check: The comparison was used to bound the integral, not to reverse its inequality. The function is positive, continuous, and decreasing on the full interval required by the test.

Worked Example: Direct Comparison for Convergence

Classify \(\sum_{n=1}^{\infty}\frac{1}{n^2+\sqrt n}\).

For each integer \(n\ge1\), both \(n^2\) and \(\sqrt n\) are positive. Hence \(n^2+\sqrt n\ge n^2>0\). Taking reciprocals reverses the inequality between these positive quantities:

0<\frac{1}{n^2+\sqrt n}\le \frac{1}{n^2}.

The comparison series \(\sum_{n=1}^{\infty}1/n^2\) is a \(p\)-series with \(p=2>1\), so it converges. The direct comparison convergence rule therefore implies

\sum_{n=1}^{\infty}\frac{1}{n^2+\sqrt n} \quad\text{converges.}

Check: The original terms are positive and no larger than the terms of the convergent comparison series. This is the required inequality direction for proving convergence.

Worked Example: Direct Comparison for Divergence

Classify \(\sum_{n=2}^{\infty}\frac{3n+1}{n^2+4n}\).

For \(n\ge2\), the denominator factors as \(n(n+4)>0\). To compare with the harmonic series, verify the proposed lower bound directly:

\frac{3n+1}{n^2+4n}\ge\frac1n \quad\Longleftrightarrow\quad n(3n+1)\ge n^2+4n \quad\Longleftrightarrow\quad 2n^2-3n\ge0.

Since \(2n^2-3n=n(2n-3)\) and \(n\ge2\), both factors are positive. Thus the inequality holds for every term in the series. The comparison series \(\sum_{n=2}^{\infty}1/n\) diverges, so the direct comparison divergence rule gives

\sum_{n=2}^{\infty}\frac{3n+1}{n^2+4n} \quad\text{diverges.}

Check: The lower bound, not an upper bound, is what transfers divergence from the harmonic series to the given series.

Worked Example: Divergence by the Integral Test

Classify \(\sum_{n=2}^{\infty}\frac{1}{n\ln n}\).

Let \(f(x)=\frac{1}{x\ln x}\) for \(x\ge2\). Since \(x>1\), both \(x\) and \(\ln x\) are positive. The function is continuous on \([2,\infty)\). Its derivative is

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

For \(x\ge2\), the numerator \(\ln x+1\) and denominator \(x^2(\ln x)^2\) are positive, so \(f'(x)<0\). Thus \(f\) is decreasing and all hypotheses of the integral test hold.

For \(R>2\), use the substitution \(u=\ln x\), so \(du=dx/x\):

\int_2^R\frac{1}{x\ln x}\,dx =\int_{\ln 2}^{\ln R}\frac{1}{u}\,du =\ln(\ln R)-\ln(\ln 2).

As \(R\to\infty\), \(\ln R\to\infty\), and therefore \(\ln(\ln R)\to\infty\). The improper integral diverges. By the integral test,

\sum_{n=2}^{\infty}\frac{1}{n\ln n} \quad\text{diverges.}

Check: The divergence comes from the unbounded integral, and the function’s positivity, continuity, and decreasing behavior have each been verified.

Choosing a Comparison Without Reversing the Logic

For a series of positive terms, first look for a simple benchmark that bounds the given terms in the useful direction. To prove convergence, an upper bound by a convergent series is useful. To prove divergence, a lower bound by a divergent series is useful.

Goal	Useful inequality	Benchmark behavior
Prove convergence	\(0\le a_n\le b_n\)	\(\sum b_n\) converges
Prove divergence	\(a_n\ge b_n\ge0\)	\(\sum b_n\) diverges
Use the integral test	\(a_n=f(n)\)	\(f\) is positive, continuous, decreasing; evaluate \(\int_N^\infty f(x)\,dx\)

A common mistake is to reverse the comparison logic. For example, from \(a_n\ge b_n\) and convergence of \(\sum b_n\), one cannot conclude that \(\sum a_n\) converges. A larger positive-term series may diverge. Likewise, \(a_n\le b_n\) and divergence of \(\sum b_n\) do not prove that \(\sum a_n\) diverges: a smaller series may converge.

For a concrete illustration, \(1/n^2\le1/n\), but \(\sum1/n^2\) converges while \(\sum1/n\) diverges. The inequality alone is not enough; the inequality and the known behavior of the benchmark must be paired in the correct way. If an inequality holds only for \(n\ge N\), that is sufficient, because the first \(N-1\) terms cannot change convergence or divergence.

Check Your Understanding

For each question, identify the test or comparison that would support the conclusion, and state the hypothesis or inequality that must be checked.

For \(\sum_{n=1}^{\infty}\frac{1}{n^2+3}\), which convergent benchmark gives a direct comparison?
To prove that a positive-term series diverges by comparison, should its terms be bounded above or below by the terms of a divergent series?
What three properties of \(f\) must be checked before applying the integral test on \([N,\infty)\)?
Does \(0\le a_n\le b_n\), together with divergence of \(\sum b_n\), establish divergence of \(\sum a_n\)? Explain the direction issue.
If an integral-test function is decreasing only for \(x\ge N\), can the test still be used for the series tail beginning at \(N\)?