Series Mastery I

From a Single Test to a Convergence Strategy

Dirichlet’s test separates a weighted series into cancellation, supplied by bounded partial sums, and decay, supplied by weights that decrease to zero. A closely related question is what happens when the unweighted series already converges, but the multiplier decreases or increases to a nonzero limit. The answer is Abel’s test: a convergent series remains convergent after multiplication by a bounded monotone sequence.

The proof uses the same central tool as Dirichlet’s test: finite summation by parts. The difference lies in what controls the tails. For Dirichlet’s test, the weights themselves become small. For Abel’s test, the partial sums of the original series over a sufficiently late block become small, while the multiplier has bounded total variation on that block. This distinction gives a useful way to choose between the two tests.

A Uniform Estimate for Monotone Multipliers

Suppose the partial sums of a series $\sum a_n$ converge. Convergence implies that finite sums taken sufficiently far out are uniformly small: given $\eta>0$, we can choose $P$ so that $ \left|\sum_{n=p}^{k}a_n\right|<\eta $ whenever $k\geq p\geq P$. The next estimate explains how that control combines with a bounded monotone multiplier.

Theorem (Uniform Tail Estimate for Monotone Multipliers): Suppose $1\leq p\leq q$, $|b_n|\leq B$ for every $n$, and $b_n$ is monotone. Define $C_k=\sum_{n=p}^{k}a_n$ for $p\leq k\leq q$. If $|C_k|\leq\eta$ for every such $k$, then $$ \left|\sum_{n=p}^{q}a_nb_n\right|\leq 3B\eta. $$

Proof. Apply the Finite Summation-by-Parts Identity to the block from $p$ to $q$, using the block partial sums $C_k$. Since $C_{p-1}=0$, it gives

\sum_{n=p}^{q}a_nb_n = C_qb_q+\sum_{n=p}^{q-1}C_n(b_n-b_{n+1}).

Take absolute values and use $|C_k|\leq\eta$ and $|b_q|\leq B$. Monotonicity ensures that all the differences $b_n-b_{n+1}$ have the same sign, so the sum of their absolute values is $\left|b_p-b_q\right|$. Hence

\left|\sum_{n=p}^{q}a_nb_n\right| \leq \eta B+\eta\sum_{n=p}^{q-1}|b_n-b_{n+1}| =\eta B+\eta|b_p-b_q| \leq 3B\eta.

The final inequality follows from $|b_p-b_q|\leq |b_p|+|b_q|\leq2B$. If $p=q$, the sum of differences is empty and equals zero, and the same bound holds. This proves the estimate. $\square$

The estimate is uniform in the length of the block: once its initial index is late enough to make all block partial sums small, the multiplier cannot turn that smallness into a large weighted sum. The monotonicity hypothesis matters because it lets us control the total variation of $b_n$ between the endpoints.

Abel’s Test

Theorem (Abel’s Test): Suppose $\sum_{n=1}^{\infty}a_n$ converges, and $(b_n)$ is a bounded monotone sequence of real numbers. Then $\sum_{n=1}^{\infty}a_nb_n$ converges.

Proof. Let $B\geq0$ satisfy $|b_n|\leq B$ for every $n$. If $B=0$, then every $b_n=0$, so the conclusion is immediate. Suppose $B>0$, and fix $\varepsilon>0$. Since $\sum a_n$ converges, its partial sums satisfy the Cauchy criterion. Choose $P$ so that

\left|\sum_{n=p}^{k}a_n\right|<\frac{\varepsilon}{3B} \qquad\text{whenever }k\geq p\geq P.

For any $q\geq p\geq P$, apply the Uniform Tail Estimate for Monotone Multipliers with $\eta=\varepsilon/(3B)$. The block partial sums $C_k=\sum_{n=p}^{k}a_n$ satisfy its hypothesis, and therefore

\left|\sum_{n=p}^{q}a_nb_n\right| \leq 3B\eta =\varepsilon.

Thus every sufficiently late finite block of the product series has absolute value at most $\varepsilon$. To obtain the strict inequality in the Cauchy criterion, begin instead with $\eta=\varepsilon/(6B)$; the same estimate then gives a bound of $\varepsilon/2<\varepsilon$. The Cauchy Criterion for Series proves that $\sum a_nb_n$ converges. $\square$

The proof does not require $b_n\to0$. A bounded monotone sequence may approach any finite limit, including a nonzero one. Nor must $b_n$ be nonnegative: the proof uses boundedness and monotonicity, not a fixed sign.

Worked Examples: Applying Abel’s Test

Worked Example: A Conditional Series with a Multiplier Approaching One

Consider

\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\left(1-\frac{1}{n+1}\right).

Set $a_n=(-1)^{n-1}/n$ and $b_n=1-1/(n+1)=n/(n+1)$. The series $\sum a_n$ converges by the Alternating Series Test: $1/n$ is nonnegative, decreases, and tends to zero. The multiplier is bounded, since $0<n/(n+1)<1$, and is increasing because

b_{n+1}-b_n = \frac{n+1}{n+2}-\frac{n}{n+1} = \frac{1}{(n+1)(n+2)} >0.

Abel’s test therefore proves convergence. In fact, the product term simplifies to $(-1)^{n-1}/(n+1)$, but the test establishes convergence directly from the original series and the multiplier. The convergence is not absolute: the absolute product term is $1/(n+1)$, and

\lim_{n\to\infty} \frac{1/(n+1)}{1/n} = \lim_{n\to\infty}\frac{n}{n+1} =1.

The Limit Comparison Test with the divergent harmonic series shows that the absolute-value series diverges. This example demonstrates that Abel’s test can preserve conditional convergence even when the multiplier tends to one rather than zero.

Worked Example: A Decreasing Multiplier with a Nonzero Limit

Consider

\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt n}\left(1+\frac{1}{n+1}\right).

Again, $a_n=(-1)^{n-1}/\sqrt n$ gives a convergent series by the Alternating Series Test: $1/\sqrt n$ decreases to zero. Set $b_n=1+1/(n+1)$. This sequence is positive and bounded above by $3/2$, since $n\geq1$. It is decreasing because

b_n-b_{n+1} = \frac{1}{n+1}-\frac{1}{n+2} = \frac{1}{(n+1)(n+2)} >0.

Abel’s test proves convergence of the product series. Its absolute-value series diverges: $b_n\geq1$, so $b_n/\sqrt n\geq1/\sqrt n$, and the $p$-series with $p=1/2$ diverges. Thus the product series is conditionally convergent. Notice that the multiplier tends to $1$; asking it to tend to zero would impose a stronger condition than Abel’s test needs.

Worked Example: Why Boundedness Alone Is Not Enough

Let $a_n=(-1)^{n-1}/n$ and $c_n=(-1)^{n-1}$. The series $\sum a_n$ converges by the Alternating Series Test, and $(c_n)$ is bounded because $|c_n|=1$. But $c_n$ is not monotone: it alternates between $1$ and $-1$. Multiplication gives

a_nc_n = \frac{(-1)^{n-1}}{n}(-1)^{n-1} = \frac{1}{n}.

The product series is therefore the harmonic series, which diverges by the $p$-Series Convergence Criterion with $p=1$. This does not contradict Abel’s test, because the multiplier is not monotone. It shows why the theorem cannot be applied using boundedness alone.

Choosing Between Abel’s and Dirichlet’s Tests

The two tests use the same kind of finite summation-by-parts estimate, but their hypotheses organize the information differently. In Dirichlet’s test, the partial sums of $a_n$ are bounded, and the multiplier decreases to zero. In Abel’s test, the series $\sum a_n$ already converges, and the multiplier is bounded and monotone. These are different sufficient conditions, not interchangeable descriptions of one hypothesis.

Test	Information about the original terms	Information about the multiplier	Conclusion
Dirichlet’s test	Partial sums are bounded	Nonnegative, nonincreasing, and tends to zero	The product series converges
Abel’s test	The original series converges	Bounded and monotone	The product series converges

A practical strategy is to inspect the two factors separately. If the original series converges and the multiplier is bounded and monotone, Abel’s test is available, even if the multiplier does not approach zero. If the terms $a_n$ have bounded partial sums but their series does not converge, Dirichlet’s test may still apply when the multiplier decreases to zero. If neither pattern fits, another method may be needed; a familiar-looking oscillation or decay is not by itself a proof.

Check the original series.
Can you prove that $\sum a_n$ converges? If so, proceed to the multiplier.

Check the multiplier.
If $b_n$ is both bounded and monotone, Abel’s test proves convergence of $\sum a_nb_n$.

If the original series is not known to converge, check its partial sums.
If they are bounded and $b_n$ is nonnegative, nonincreasing, and tends to zero, Dirichlet’s test applies.

Separate convergence from absolute convergence.
Neither test alone establishes convergence of $\sum |a_nb_n|$; analyze that series separately.

A final common pitfall is to infer absolute convergence from a successful Abel or Dirichlet test. Both proofs control signed finite sums, where cancellation is essential. The first worked example showed that an Abel-tested series can converge while its absolute-value series diverges. Whenever the conclusion needed is absolute convergence, apply an absolute-value comparison or another suitable test to $|a_nb_n|$.

Takeaway: Abel’s test applies when the original series converges and the multiplier is bounded and monotone. The uniform tail estimate explains why: small partial sums on late blocks remain controlled after multiplication by a sequence with bounded variation.

Check Your Understanding

Use the hypotheses and estimates in this tutorial to answer the following questions.

Where does monotonicity enter the proof of the Uniform Tail Estimate for Monotone Multipliers?
Why does Abel’s test not require the multiplier to tend to zero?
Which test applies if the partial sums of $a_n$ are bounded, but convergence of $\sum a_n$ is not known?
For $a_n=(-1)^{n-1}/n$ and $b_n=n/(n+1)$, verify that $b_{n+1}-b_n>0$.
Why does convergence from Abel’s test not imply absolute convergence?

What You'll Learn