Comparison of Positive Series

From Telescoping to Comparing Partial Sums

In the previous tutorial, a finite telescoping identity made the partial sums explicit: cancellation left boundary terms, and their behavior determined whether the series converged. A different strategy is useful when no such cancellation appears. If the terms are nonnegative, an inequality between individual terms gives an inequality between partial sums. We can then use a series whose behavior is already known to control the one we are studying.

The key is to work first with finite sums. Infinite series are defined through limits of partial sums, so a term-by-term inequality cannot be used as an argument about infinite sums until its consequences for those partial sums are established. Nonnegativity matters: it makes the partial sums increasing and prevents later terms from undoing an earlier lower bound.

Definition: A series $\sum_{n=1}^{\infty} a_n$ is called a positive-term series here if $a_n\geq 0$ for every $n$. Its $N$th partial sum is $S_N=\sum_{n=1}^{N}a_n$. If every term is strictly positive, the series is also called a series with positive terms. The comparison arguments below allow zero terms as well.

For two positive-term series, suppose $a_n\leq b_n$ for every $n$. Adding the inequalities from $n=1$ to $N$ gives $S_N\leq T_N$, where $T_N=\sum_{n=1}^{N}b_n$. This finite-sum inequality is the basic mechanism behind comparison. It leads to two useful conclusions: a convergent upper series controls the convergence of the lower one, while a divergent lower series forces the upper one to diverge.

The Comparison Principle

Theorem (Comparison Principle for Positive-Term Series): Suppose $a_n\geq0$ and $b_n\geq0$ for every $n$.

If $a_n\leq b_n$ for every $n$ and $\sum_{n=1}^{\infty}b_n$ converges, then $\sum_{n=1}^{\infty}a_n$ converges.
If $a_n\geq b_n$ for every $n$ and $\sum_{n=1}^{\infty}b_n$ diverges, then $\sum_{n=1}^{\infty}a_n$ diverges.

Proof. Define the partial sums $S_N=\sum_{n=1}^{N}a_n$ and $T_N=\sum_{n=1}^{N}b_n$. In the first case, summing $a_n\leq b_n$ over the finite range $1\leq n\leq N$ gives

0\leq S_N\leq T_N.

Since $\sum b_n$ converges, its partial sums $T_N$ are bounded above. Thus $S_N$ is bounded above as well. Also, $S_N$ is nondecreasing because $S_{N+1}-S_N=a_{N+1}\geq0$. By the earlier Theorem (Bounded Increasing Partial Sums), $S_N$ converges, which means $\sum a_n$ converges.

For the second case, $a_n\geq b_n$ gives $S_N\geq T_N$ for every $N$. The partial sums $T_N$ are nondecreasing and do not converge to a finite value, by the assumed divergence. A nondecreasing sequence of real numbers that is unbounded above cannot converge to a finite real number. Since $S_N\geq T_N$, the sequence $S_N$ is unbounded above as well. Therefore $\sum a_n$ diverges. $\square$

Notice the direction of the inequalities. To prove convergence, find a convergent series above the series of interest. To prove divergence, find a divergent series below it. Reversing either inequality does not give the stated conclusion.

Worked Example: Convergence by a Geometric Upper Bound

Consider $\sum_{n=1}^{\infty}1/(3^n+2)$. Each term is positive, and $3^n+2\geq3^n>0$. Taking reciprocals of positive quantities reverses the inequality, so

0<\frac{1}{3^n+2}\leq\frac{1}{3^n}=\left(\frac13\right)^n.

The upper series is geometric with ratio $1/3$, whose absolute value is less than $1$. By the Convergence Criterion for a Geometric Series, it converges; its sum, by the Sum of a Convergent Geometric Series, is

\sum_{n=1}^{\infty}\left(\frac13\right)^n =\frac{1/3}{1-1/3} =\frac12.

The Comparison Principle therefore shows that $\sum_{n=1}^{\infty}1/(3^n+2)$ converges. The argument does not claim that its sum is $1/2$; the upper bound only establishes convergence and a bound on the sum.

Comparison with a Divergent Lower Series

A lower bound can be just as useful when the goal is to prove divergence. For example, the harmonic series $\sum_{n=1}^{\infty}1/n$ diverges. One way to see this is to group terms in blocks whose lengths double: the terms with indices $2^{k-1}+1$ through $2^k$ each have size at least $1/2^k$, and there are $2^{k-1}$ terms in that block. Its sum is therefore at least $1/2$. Infinitely many such blocks prevent the partial sums from being bounded.

Worked Example: A Harmonic Lower Bound Forces Divergence

Consider the positive-term series $\sum_{n=1}^{\infty}(n+2)/(n+1)^2$. For every $n\geq1$, its numerator $n+2$ is greater than $n+1$, and its denominator $(n+1)^2$ is positive. Hence

\frac{n+2}{(n+1)^2}>\frac{n+1}{(n+1)^2} =\frac{1}{n+1}.

The lower series $\sum_{n=1}^{\infty}1/(n+1)$ diverges: its $N$th partial sum is the harmonic partial sum $1+1/2+\cdots+1/(N+1)$ with its first term removed, and removing one finite term cannot turn an unbounded sequence of partial sums into a bounded one. By the divergence part of the Comparison Principle,

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

The comparison is strict for each term, but strictness is not what drives the conclusion. The essential fact is that the terms are at least as large as the terms of a divergent positive-term series.

Tail Bounds and What Convergence Gives

Comparison can also estimate how much of a convergent series remains after a partial sum. Suppose $0\leq a_n\leq b_n$, and both series converge, with sums $A$ and $B$. For $M>N$, summing over the tail range gives

0\leq\sum_{n=N+1}^{M}a_n \leq\sum_{n=N+1}^{M}b_n.

As $M$ tends to infinity, the two finite tail sums tend to the corresponding remainders. Thus the comparison also bounds the infinite tail. This conclusion is often valuable when a numerical approximation is needed: it provides an explicit error estimate for replacing a series by a partial sum.

Theorem (Tail Bound from a Convergent Comparison): Suppose $0\leq a_n\leq b_n$ for every $n$, and suppose $\sum a_n$ and $\sum b_n$ converge. Write $R_N^a=\sum_{n=N+1}^{\infty}a_n$ and $R_N^b=\sum_{n=N+1}^{\infty}b_n$. Then $$ 0\leq R_N^a\leq R_N^b $$ for every nonnegative integer $N$.

Proof. For each $M>N$, termwise comparison over the finite range $N+1\leq n\leq M$ gives

0\leq\sum_{n=N+1}^{M}a_n \leq\sum_{n=N+1}^{M}b_n.

By convergence of the two series, the left and right finite tail sums tend respectively to $R_N^a$ and $R_N^b$ as $M$ tends to infinity. Taking limits preserves the inequalities, so $0\leq R_N^a\leq R_N^b$. $\square$

Worked Example: An Explicit Remainder Estimate

For $\sum_{n=1}^{\infty}1/(4^n+1)$, the earlier inequality $4^n+1\geq4^n$ gives \[ 0<\frac{1}{4^n+1}\leq\frac{1}{4^n}. \] The upper geometric series converges, so the series of interest converges as well. The tail of the upper series can be evaluated exactly. For $N\geq0$,

\sum_{n=N+1}^{\infty}\frac{1}{4^n} =\frac{(1/4)^{N+1}}{1-1/4} =\frac{1}{3\cdot4^N}.

The Tail Bound from a Convergent Comparison therefore gives \[ 0\leq\sum_{n=N+1}^{\infty}\frac{1}{4^n+1} \leq\frac{1}{3\cdot4^N}. \] For instance, after summing through $n=3$, the remainder is at most $1/(3\cdot4^3)=1/192$. This is a certified error bound, not merely an indication that the remainder tends to zero.

Why a One-Sided Comparison Is Not an Equivalence

A common mistake is to treat comparison as though it gave an “if and only if” test. It does not. If $0\leq a_n\leq b_n$ and $\sum b_n$ diverges, nothing follows in general about $\sum a_n$: the upper series may diverge even when the smaller series converges. Likewise, if $\sum a_n$ converges, a larger series might converge or diverge.

Worked Example: A Convergent Series Below a Divergent One

For every $n\geq1$, $2^n\geq n$. This follows by induction: it holds for $n=1$, and if $2^n\geq n$, then $2^{n+1}\geq2n\geq n+1$. Since these quantities are positive, taking reciprocals gives

0<\frac{1}{2^n}\leq\frac{1}{n}.

The series $\sum_{n=1}^{\infty}1/2^n$ converges by the geometric-series criterion, while the harmonic series $\sum_{n=1}^{\infty}1/n$ diverges. Thus a convergent series can lie below a divergent series term by term. The comparison conclusion for convergence requires the upper series to converge; a divergent upper bound is not enough to decide the lower series.

Positivity is also essential to this method. For nonnegative terms, partial sums are nondecreasing, so boundedness guarantees convergence by the earlier Theorem (Bounded Increasing Partial Sums). With signed terms, later negative terms can cancel earlier positive terms, and termwise inequalities do not give the same control over partial sums. For series that are not positive-term series, a separate argument is needed.

Check the signs.
Verify that the terms in the series being tested and the comparison series are nonnegative.

Choose the direction.
For convergence, seek a convergent series above the terms of interest. For divergence, seek a divergent series below them.

Verify the inequality.
Check the termwise bound for every index in the stated range, including any initial terms.

Apply the conclusion carefully.
A one-sided bound proves the relevant implication; it does not generally determine the behavior in the opposite direction.

Takeaway: For positive-term series, inequalities between terms pass to finite partial sums. A convergent upper series proves convergence, and a divergent lower series proves divergence. When the comparison series converges, its tail also supplies a bound for the remainder.

Check Your Understanding

Use finite partial sums and the direction of each inequality to answer the following questions.

If $0\leq a_n\leq b_n$ and $\sum b_n$ converges, which series is guaranteed to converge?
If $a_n\geq b_n\geq0$ and $\sum b_n$ diverges, what conclusion follows about $\sum a_n$?
Why does $0\leq a_n\leq b_n$ not prove that $\sum a_n$ diverges when $\sum b_n$ diverges?
If both series converge and $0\leq a_n\leq b_n$, how are their remainders after the $N$th partial sum related?
What role does nonnegativity play in using bounded partial sums to prove convergence?

What You'll Learn