Positive-Term Series and Order

Why Compare Terms?

A series can be difficult to sum directly even when its terms resemble those of a familiar series. A useful first question is whether its terms are larger or smaller than the terms of a series whose behavior is already known. For nonnegative terms, an inequality between the terms also orders every pair of corresponding partial sums. That simple fact lets us transfer convergence or divergence in specific directions.

The direction matters. If a series with larger terms converges, then the smaller one must converge too. If a series with smaller terms diverges, then the larger one must diverge too. The converse conclusions do not follow just from the inequality.

Nonnegative Series and Partial Sums

Definition: A series $\sum_{n=1}^{\infty} a_n$ is a nonnegative-term series if $a_n\ge 0$ for every $n$. Its $N$th partial sum is $S_N=\sum_{n=1}^{N}a_n$. The series converges if the sequence of partial sums has a finite limit as $N\to\infty$; otherwise it diverges.

Because each term is nonnegative, $S_{N+1}=S_N+a_{N+1}\ge S_N$. Thus the partial sums never decrease. A nondecreasing sequence of real numbers converges to a finite limit exactly when it is bounded above. Consequently, a nonnegative-term series either converges to a finite sum or its partial sums grow without bound.

For completeness, if the partial sums are bounded above, let $L$ be their supremum. For every $\varepsilon>0$, the definition of supremum gives some $S_M>L-\varepsilon$. Since the partial sums are nondecreasing, for every $N\ge M$ we have $L-\varepsilon<S_M\le S_N\le L$. Therefore $S_N\to L$. If the partial sums are not bounded above, they cannot converge to a finite limit.

Termwise Order and Partial-Sum Order

Key fact: If $0\le a_n\le b_n$ for every $n$, then for each positive integer $N$, $$ 0\le \sum_{n=1}^{N}a_n\le \sum_{n=1}^{N}b_n. $$

This follows by adding the inequalities $a_1\le b_1,\ldots,a_N\le b_N$. No limit operation is needed to obtain the partial-sum inequality: it holds separately for every finite $N$.

Write $S_N=\sum_{n=1}^{N}a_n$ and $T_N=\sum_{n=1}^{N}b_n$. If $\sum b_n$ converges, then its partial sums $T_N$ are bounded above by their finite limit. Since $0\le S_N\le T_N$, the partial sums $S_N$ are bounded too; their nondecreasing property then gives convergence. In the other direction, if $\sum a_n$ diverges, its nondecreasing partial sums are unbounded. Since $T_N\ge S_N$, the partial sums $T_N$ are unbounded as well, so $\sum b_n$ diverges.

Comparison principle: For nonnegative terms with $a_n\le b_n$:

If $\sum b_n$ converges, then $\sum a_n$ converges.
If $\sum a_n$ diverges, then $\sum b_n$ diverges.

These are the two conclusions licensed by this order. In particular, knowing that the smaller series converges says nothing by itself about whether the larger series converges. Likewise, knowing that the larger series diverges does not by itself determine the smaller one.

Worked Comparisons

Worked Example: A Smaller Series Than a Convergent p-Series

Consider $\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$. Compare its terms with those of $\sum_{n=1}^{\infty}\frac{1}{n^2}$.

For $n\ge1$, $n(n+1)=n^2+n\ge n^2$. Both denominators are positive, so taking reciprocals reverses the inequality:

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

The p-series $\sum_{n=1}^{\infty}\frac{1}{n^2}$ converges because its exponent is $2>1$. By comparison, $\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$ converges.

As a check of the terms and the conclusion, the identity $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ gives the finite sum

\sum_{n=1}^{N}\frac{1}{n(n+1)} =\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\cdots+ \left(\frac1N-\frac{1}{N+1}\right) =1-\frac{1}{N+1}.

As $N\to\infty$, this partial sum tends to $1$, confirming convergence and giving the sum.

Worked Example: A Larger Series Than a Divergent Series

Consider $\sum_{n=1}^{\infty}\frac{1}{\sqrt n}$. For each $n\ge1$, $\sqrt n\le n$, because squaring both nonnegative sides gives $n\le n^2$. Taking reciprocals yields

0<\frac{1}{n}\le\frac{1}{\sqrt n}.

The harmonic series $\sum_{n=1}^{\infty}\frac{1}{n}$ diverges. Its partial sums are therefore unbounded. The partial sums of $\sum_{n=1}^{\infty}\frac{1}{\sqrt n}$ are at least as large term by term, so they too are unbounded. Hence $\sum_{n=1}^{\infty}\frac{1}{\sqrt n}$ diverges.

For a numerical check of the inequality, at $n=4$ the compared terms are $\frac14\le\frac12$. The direction agrees with the general conclusion: divergence of the smaller series forces divergence of the larger one.

Worked Example: Why the Reverse Conclusions Fail

For every $n\ge1$, $\frac{1}{n^2}\le\frac{1}{n}$. It would be incorrect to conclude that the larger series must converge because the smaller series converges. The comparison principle permits that conclusion only when the larger series is known to converge.

Indeed, $\sum_{n=1}^{\infty}\frac{1}{n^2}$ converges, while $\sum_{n=1}^{\infty}\frac{1}{n}$ diverges. For example, their first three partial sums are

\sum_{n=1}^{3}\frac{1}{n^2} =1+\frac14+\frac19 =\frac{36}{36}+\frac{9}{36}+\frac{4}{36} =\frac{49}{36}, $$ $$ \sum_{n=1}^{3}\frac{1}{n} =1+\frac12+\frac13 =\frac{6}{6}+\frac{3}{6}+\frac{2}{6} =\frac{11}{6}.

The finite sums satisfy $\frac{49}{36}\le\frac{11}{6}$, as required by the termwise inequality. The established classifications also confirm the point: the smaller series converges, but the larger one does not.

Comparisons That Hold Eventually

Sometimes an inequality is true only from some index $N_0$ onward. This is enough for convergence comparisons. The first $N_0-1$ terms contribute only a fixed finite amount to the partial sums; changing or adding finitely many terms cannot turn a finite limit into an infinite one, or an unbounded sequence of partial sums into a bounded one.

Worked Example: A Comparison After the First Term

Define $c_1=100$ and $c_n=\frac{1}{n(n+1)}$ for $n\ge2$. For $n\ge2$, the same denominator comparison gives

0<c_n=\frac{1}{n(n+1)}\le\frac{1}{n^2}.

The p-series tail $\sum_{n=2}^{\infty}\frac{1}{n^2}$ converges, so comparison shows that $\sum_{n=2}^{\infty}c_n$ converges. The initial term $c_1=100$ is a single finite number and does not affect convergence.

We can check the result directly. For $N\ge2$,

\sum_{n=1}^{N}c_n =100+\sum_{n=2}^{N}\left(\frac1n-\frac{1}{n+1}\right) =100+\frac12-\frac{1}{N+1}.

This tends to $100+\frac12=\frac{201}{2}$. The first term is large, but finite; it changes the sum, not whether the series converges.

A Common Direction-of-Inequality Mistake

A frequent error is to start with $a_n\le b_n$, observe that $\sum a_n$ converges, and claim that $\sum b_n$ must converge too. The inequality gives $S_N\le T_N$, so a bound on the smaller partial sums $S_N$ is not an upper bound on the larger partial sums $T_N$. Example 3 shows the failure concretely: $\sum 1/n^2$ converges, but the larger harmonic series diverges.

Write the termwise inequality.
Identify which series has the smaller terms and which has the larger terms.

Check the known series.
Use a series whose convergence or divergence has already been established.

Transfer only in the valid direction.
A convergent larger series bounds the smaller one; a divergent smaller series forces the larger one to diverge.

Why Order Is Useful

Termwise comparison turns a difficult infinite sum into an order question that can be checked at each index. It is especially useful when a complicated expression has terms that can be bounded by a familiar benchmark. The comparison itself does not require the terms to have a closed-form sum; it requires nonnegative terms, a valid inequality, and a benchmark whose behavior is known.

Always verify the inequality before applying the conclusion. When taking reciprocals, for instance, the direction reverses only when both quantities are positive. Also check whether the inequality holds for every index or merely eventually; a finite initial segment is harmless, but the remaining terms must satisfy the comparison consistently.

Check Your Understanding

For each question, identify the direction of the inequality and state exactly what conclusion it permits.

If $0\le a_n\le b_n$ and $\sum b_n$ converges, what can be concluded about $\sum a_n$?
If $0\le a_n\le b_n$ and $\sum a_n$ diverges, what can be concluded about $\sum b_n$?
Why does convergence of $\sum a_n$ not, by itself, imply convergence of $\sum b_n$ under the same inequality?
Does a comparison need to hold for the first term if it holds for every term from some fixed index onward? Explain.
If a nonnegative series has bounded partial sums, why must it converge?

What You'll Learn