Absolute and Conditional Convergence

Some series converge only because their positive and negative terms cancel, like $1 - \tfrac{1}{2} + \tfrac{1}{3} - \dots$ Others would converge even if every term were made positive, like $1 - \tfrac{1}{4} + \tfrac{1}{9} - \dots$ This page names that difference (conditional versus absolute convergence), shows how to classify any series into one of three categories, and pulls all the convergence tests of this unit together into one strategy.

Key ideas

Two kinds of convergence

For a series $\sum a_n$ whose terms may be positive or negative:

• $\sum a_n$ converges absolutely if the series of absolute values $\sum \lvert a_n \rvert$ converges.
• $\sum a_n$ converges conditionally if $\sum a_n$ converges but $\sum \lvert a_n \rvert$ diverges.

Every series ends up in exactly one of three boxes: absolutely convergent, conditionally convergent, or divergent.

Absolute convergence implies convergence

$$\text{If } \sum \lvert a_n \rvert \text{ converges, then } \sum a_n \text{ converges.}$$

Making some terms negative can only create cancelling, never more growth. This is a powerful fact: it lets you use the tests for positive series (comparison, integral, p-series) on series with mixed signs, by testing $\sum \lvert a_n \rvert$.

The converse is false. The alternating harmonic series converges, but its absolute values form the harmonic series, which diverges.

How to classify a series

1. Check the terms first. If they don’t approach $0$, the series diverges by the nth term test. It saves time.
2. Test $\sum \lvert a_n \rvert$ (using p-series, comparison, the integral test, or the ratio test). If it converges, the series converges absolutely. Done.
3. If $\sum \lvert a_n \rvert$ diverges, test $\sum a_n$ itself. For an alternating series, use the alternating series test. If it converges, the series converges conditionally.

When the ratio test gives $L \lt 1$, the series converges absolutely, since the test uses $\lvert a_n \rvert$. When it gives $L \gt 1$, the series diverges.

Strategy: which test to use

If the series looks like…	Try
terms clearly don’t go to $0$	$n$th term test (diverges)
$\sum a r^n$ ($n$ only in exponents)	geometric series: converges exactly when $\lvert r \rvert \lt 1$
$\sum \frac{1}{n^p}$, or roots and powers of $n$	p-series: converges exactly when $p \gt 1$
a rational or algebraic expression in $n$	direct or limit comparison with a p-series
a difference that cancels, like $\frac{1}{n} - \frac{1}{n+1}$	telescoping: find $S_n$
factorials, or $n$ in an exponent mixed with other factors	ratio test
$(-1)^n$ or $\cos(n\pi)$ times a positive term	test absolute values first, then the alternating series test
$f(n)$ where $f$ is easy to integrate (like $\frac{1}{n \ln n}$)	integral test
$\sin n$ or $\cos n$ in the numerator	compare $\lvert a_n \rvert$ using $\lvert \sin n \rvert \le 1$

On the AP exam, a “converges absolutely, converges conditionally, or diverges?” question needs two pieces of work for a conditionally convergent series: one test showing $\sum \lvert a_n \rvert$ diverges and another showing $\sum a_n$ converges. Name each test and check its conditions.

Worked examples

Example 1: Absolute or conditional?

Classify each series as absolutely convergent, conditionally convergent, or divergent.

• (a) $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$
• (b) $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$

Solution.

(a) The absolute values are $\sum \frac{1}{n^2}$, a p-series with $p = 2 \gt 1$, which converges. So the series converges absolutely.

(b) The absolute values are $\sum \frac{1}{n}$, the harmonic series, which diverges. So it doesn’t converge absolutely. But the series itself converges by the alternating series test ($\frac{1}{n}$ decreases to $0$). So it converges conditionally.

Example 2: Comparison, then the alternating series test

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

Solution. Absolute values: $\sum \dfrac{n}{n^2 + 1}$. Compare with $\frac{1}{n}$:

$$\lim_{n \to \infty} \frac{n}{n^2 + 1} \cdot n = \lim_{n \to \infty} \frac{n^2}{n^2 + 1} = 1$$

Since $0 \lt 1 \lt \infty$ and the harmonic series diverges, $\sum \lvert a_n \rvert$ diverges by the limit comparison test.

The series itself: it alternates with $b_n = \dfrac{n}{n^2 + 1}$, which is decreasing for $n \ge 1$ (its derivative $\frac{1 - x^2}{(x^2 + 1)^2}$ is $\le 0$) and has limit $0$. So it converges by the alternating series test.

Conclusion: the series converges conditionally.

Example 3: Mixed signs that don’t alternate

Does $\displaystyle\sum_{n=1}^{\infty} \frac{\sin n}{n^2}$ converge? (The angle $n$ is in radians.)

Solution. The signs of $\sin n$ follow no regular pattern ($\sin 1, \sin 2, \sin 3$ are positive, $\sin 4, \sin 5, \sin 6$ are negative, …), so this isn’t an alternating series. Test the absolute values instead:

$$0 \le \left\lvert \frac{\sin n}{n^2} \right\rvert \le \frac{1}{n^2}$$

since $\lvert \sin n \rvert \le 1$. The p-series $\sum \frac{1}{n^2}$ converges, so $\sum \left\lvert \frac{\sin n}{n^2} \right\rvert$ converges by direct comparison. The series converges absolutely, and therefore converges.

Example 4: Using the strategy

Classify each series.

• (a) $\displaystyle\sum_{n=1}^{\infty} \frac{(-2)^n}{n^3}$
• (b) $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n \arctan n}{n^2}$

Solution.

(a) There’s an exponential, so try the ratio test:

$$\left\lvert \frac{a_{n+1}}{a_n} \right\rvert = \frac{2^{n+1}}{(n+1)^3} \cdot \frac{n^3}{2^n} = 2\left(\frac{n}{n+1}\right)^3 \to 2$$

$L = 2 \gt 1$, so the series diverges.

(b) Test the absolute values. Since $0 \lt \arctan n \lt \frac{\pi}{2}$ for $n \ge 1$,

$$\left\lvert \frac{(-1)^n \arctan n}{n^2} \right\rvert \lt \frac{\pi/2}{n^2}$$

and $\sum \frac{\pi/2}{n^2}$ is a constant times a convergent p-series. By direct comparison, $\sum \lvert a_n \rvert$ converges, so the series converges absolutely.

Common mistakes

Thinking the alternating series test shows absolute convergence. It only shows that $\sum a_n$ converges. To decide “absolutely” or “conditionally,” you must also test $\sum \lvert a_n \rvert$ separately.

Saying “conditionally convergent” means “doesn’t really converge.” A conditionally convergent series does converge, to a definite sum. It’s just that the convergence depends on the signs.

Concluding divergence because the absolute values diverge. If $\sum \lvert a_n \rvert$ diverges, the series might still converge conditionally (like the alternating harmonic series). You have to test $\sum a_n$ too.

Using the alternating series test on a series that doesn’t alternate. For $\sum \frac{\sin n}{n^2}$, the signs are irregular. Use absolute convergence with a comparison instead.

Doing only half the work for “conditionally.” On the AP exam, “converges conditionally” earns full credit only with both arguments: why $\sum \lvert a_n \rvert$ diverges, and why $\sum a_n$ converges.

Practice

1. (Warm-up) Classify $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n^3}$.

Solution

$\sum \lvert a_n \rvert = \sum \frac{1}{n^3}$, a p-series with $p = 3 \gt 1$, converges. So the series converges absolutely.

2. (Warm-up) Classify $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt[3]{n}}$.

Solution

$\sum \lvert a_n \rvert = \sum \frac{1}{n^{1/3}}$, a p-series with $p = \tfrac{1}{3} \le 1$, diverges. The series itself alternates with $b_n = \frac{1}{n^{1/3}}$, which decreases to $0$, so it converges by the alternating series test. The series converges conditionally.

3. (Warm-up) True or false? Give a reason or counterexample.

• (a) If $\sum \lvert a_n \rvert$ converges, then $\sum a_n$ converges.
• (b) If $\sum a_n$ converges, then $\sum \lvert a_n \rvert$ converges.
• (c) If $\sum \lvert a_n \rvert$ diverges, then $\sum a_n$ diverges.

Solution

(a) True: absolute convergence implies convergence.

(b) False: $\sum \frac{(-1)^{n+1}}{n}$ converges, but $\sum \frac{1}{n}$ diverges.

(c) False: the same example. $\sum \frac{1}{n}$ diverges, but $\sum \frac{(-1)^{n+1}}{n}$ converges.

4. (Core) Classify $\displaystyle\sum_{n=1}^{\infty} \frac{\cos n}{n^2 + 1}$.

Solution

The signs of $\cos n$ are irregular, so test absolute values. Since $\lvert \cos n \rvert \le 1$ and $n^2 + 1 \gt n^2$,

$$\left\lvert \frac{\cos n}{n^2 + 1} \right\rvert \le \frac{1}{n^2 + 1} \lt \frac{1}{n^2}$$

$\sum \frac{1}{n^2}$ converges, so by direct comparison $\sum \lvert a_n \rvert$ converges. The series converges absolutely.

5. (Core) Classify $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n (n + 1)}{n^2}$.

Solution

Absolute values: $\dfrac{n + 1}{n^2} = \dfrac{1}{n} + \dfrac{1}{n^2} \gt \dfrac{1}{n}$, and the harmonic series diverges, so $\sum \lvert a_n \rvert$ diverges by direct comparison.

The series itself: $b_n = \dfrac{1}{n} + \dfrac{1}{n^2}$ is a sum of two decreasing sequences, so it’s decreasing, and $b_n \to 0$. It converges by the alternating series test.

The series converges conditionally.

6. (Core) Classify $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n n^2}{3^n}$.

Solution

Ratio test:

$$\left\lvert \frac{a_{n+1}}{a_n} \right\rvert = \frac{(n+1)^2}{3^{n+1}} \cdot \frac{3^n}{n^2} = \frac{1}{3}\left(\frac{n+1}{n}\right)^2 \to \frac{1}{3}$$

$L = \tfrac{1}{3} \lt 1$, so the series converges absolutely.

7. (Core) Classify $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n \sqrt{n}}{n + 1}$.

Solution

Absolute values: $\dfrac{\sqrt{n}}{n + 1}$ behaves like $\dfrac{\sqrt{n}}{n} = \dfrac{1}{\sqrt{n}}$. Limit comparison:

$$\lim_{n \to \infty} \frac{\sqrt{n}}{n + 1} \cdot \sqrt{n} = \lim_{n \to \infty} \frac{n}{n + 1} = 1$$

$\sum \frac{1}{\sqrt{n}}$ diverges ($p = \tfrac{1}{2}$), so $\sum \lvert a_n \rvert$ diverges.

The series itself: let $f(x) = \dfrac{\sqrt{x}}{x + 1}$. By the quotient rule,

$$f'(x) = \frac{\frac{1}{2\sqrt{x}}(x + 1) - \sqrt{x}}{(x + 1)^2} = \frac{1 - x}{2\sqrt{x}\,(x + 1)^2} \le 0 \quad\text{for } x \ge 1$$

so $b_n$ is decreasing, and $\displaystyle\lim_{n \to \infty} \frac{\sqrt{n}}{n + 1} = 0$. It converges by the alternating series test.

The series converges conditionally.

8. (Challenge) For which values of $p$ is $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n^p}$ absolutely convergent, conditionally convergent, or divergent?

Solution

• $p \gt 1$: $\sum \frac{1}{n^p}$ converges, so the series converges absolutely.
• $0 \lt p \le 1$: $\sum \frac{1}{n^p}$ diverges, but $\frac{1}{n^p}$ decreases to $0$, so the alternating series test gives convergence. The series converges conditionally.
• $p \le 0$: $\lvert a_n \rvert = n^{-p} \ge 1$, so the terms don’t approach $0$. The series diverges by the $n$th term test.

9. (Challenge) Classify $\displaystyle\sum_{n=2}^{\infty} \frac{(-1)^n}{n \ln n}$.

Solution

Absolute values: $\sum_{n=2}^{\infty} \frac{1}{n \ln n}$. The function $f(x) = \frac{1}{x \ln x}$ is positive, continuous, and decreasing for $x \ge 2$. With $u = \ln x$:

$$\int_2^{\infty} \frac{1}{x \ln x}\,dx = \lim_{b \to \infty} \big(\ln(\ln b) - \ln(\ln 2)\big) = \infty$$

so $\sum \lvert a_n \rvert$ diverges by the integral test.

The series itself: $b_n = \frac{1}{n \ln n}$ is decreasing (the denominator is a product of positive increasing factors) and $b_n \to 0$. It converges by the alternating series test.

The series converges conditionally.