§15 Alternating Series and Integral Tests

Sometimes we can decide whether a series converges by comparing its partial sums with familiar integrals.

Why this matters

The tests of §14 give no information about the most important comparison series of all, $\sum\frac1{n^p}$, since their ratios tend to $1$. This section supplies two new tools: comparison with an integral, which settles the $p$-series, and the Alternating Series Theorem, which shows that a series can converge without converging absolutely.

Example 1.

We show $\sum \frac1n = +\infty$. Draw $f(x) = \frac1x$ and, over each interval $[k, k+1]$, a rectangle of height $\frac1k$. These rectangles lie above the curve, so for $n \ge 1$

$$\sum_{k=1}^n \frac1k = \text{sum of the areas of the first } n \text{ rectangles} \ge \int_1^{n+1} \frac1x\,dx = \log_e(n+1).$$

Since $\lim_{n\to\infty} \log_e(n+1) = +\infty$, we conclude $\sum_{n=1}^\infty \frac1n = +\infty$.

The series diverges very slowly. In fact $\sum_{n=1}^N \frac1n$ is approximately $\log_e N + 0.5772$, so for $N = 1{,}000$ the sum is about $7.485$, and for $N = 1{,}000{,}000$ it is only about $14.393$. Another proof of divergence was given in Exercise 14.14.

Example 2.

We show $\sum \frac1{n^2}$ converges. Now put the rectangle of height $\frac1{k^2}$ over $[k-1, k]$, so for $k \ge 2$ it lies under the curve $f(x) = \frac1{x^2}$. Then

$$\sum_{k=1}^n \frac1{k^2} = \text{sum of the areas of the first } n \text{ rectangles} \le 1 + \int_1^n \frac1{x^2}\,dx = 2 - \frac1n < 2$$

for all $n \ge 1$. The partial sums form an increasing sequence bounded above by $2$, so $\sum_{n=1}^\infty \frac1{n^2}$ converges and its sum is at most $2$. (Its sum is actually $\frac{\pi^2}6 = 1.6449\cdots$.)

We did not simply write $\sum_{k=1}^n \frac1{k^2} \le \int_0^n \frac1{x^2}\,dx$, even though it is true, because that integral is infinite. We were after a finite upper bound for the partial sums.

The harmonic series $\sum 1/n$ diverges, but its terms shrink only slightly slower than those of $\sum 1/n^{1.01}$. Does $\sum 1/n^{1.01}$ converge?

YesNo

Try it: rectangles against the curve

Set $p = 1$ with rectangles above to see Example 1, and $p = 2$ with rectangles below to see Example 2. Then move $p$ past $1$ and watch which bound stays finite.

Exponent p1Terms n8

s₈ = 2.7179 ≥ ∫₁⁹ x⁻¹ dx = 2.1972. That integral tends to +∞ as n grows, so the series diverges.

The techniques just illustrated prove the following theorem.

15.1 Theorem.

$\sum \dfrac1{n^p}$ converges if and only if $p > 1$.

Used in Exercise 15.5

Proof

Supply your own picture. If $p > 1$, then

$$\sum_{k=1}^n \frac1{k^p} \le 1 + \int_1^n \frac1{x^p}\,dx = 1 + \frac1{p-1}\left[1 - \frac1{n^{p-1}}\right] < 1 + \frac1{p-1} = \frac{p}{p-1}.$$

Consequently $\sum_{n=1}^\infty \frac1{n^p} \le \frac{p}{p-1} < +\infty$.

Suppose $0 < p \le 1$. Then $\frac1n \le \frac1{n^p}$ for all $n$. Since $\sum \frac1n$ diverges, $\sum \frac1{n^p}$ diverges by the Comparison Test.

∎

Intuition

For a decreasing $f$, the area $\int_k^{k+1} f(x)\,dx$ lies between $f(k+1)$ and $f(k)$. Adding these up, the partial sums and the integrals can never drift apart by more than the first term. So the series and the integral are both finite or both infinite.

15.2 Integral Tests.

An integral test is advisable when:

1. the tests in §14 do not seem to apply;
2. the terms $a_n$ of the series $\sum a_n$ are nonnegative;
3. there is a nice decreasing function $f$ on $[1, \infty)$ with $f(n) = a_n$ for all $n$ ($f$ is decreasing if $x < y$ implies $f(x) \ge f(y)$);
4. the integral of $f$ is easy to calculate or estimate.

If $\lim_{n\to\infty} \int_1^n f(x)\,dx = +\infty$, the series diverges just as in Example 1. If $\lim_{n\to\infty} \int_1^n f(x)\,dx < +\infty$, the series converges just as in Example 2. The interested reader may formulate and prove the general result (Exercise 15.8).

Common pitfallthe decreasing hypothesis matters

Let $f(x) = \sin^2(\pi x)$. Then $f(n) = 0$ for every integer $n$, so $\sum f(n) = 0$ converges, but $\int_1^\infty f(x)\,dx = +\infty$ since $f$ has the same positive area over every interval $[k, k+1]$. Without a decreasing $f$, the values at the integers need not reflect the area under the curve at all.

Alternating series

The next result shows that series like $\sum \frac{(-1)^n}{\sqrt n}$ converge even though they do not converge absolutely (see Example 9 of §14).

15.3 Alternating Series Theorem.

If $a_1 \ge a_2 \ge \cdots \ge a_n \ge \cdots \ge 0$ and $\lim a_n = 0$, then the alternating series $\sum (-1)^{n+1} a_n$ converges. Moreover, the partial sums $s_n = \sum_{k=1}^n (-1)^{k+1} a_k$ satisfy $|s - s_n| \le a_n$ for all $n$.

Used in §14

The series $\sum (-1)^n a_n$ is called alternating because the signs of its terms alternate between $+$ and $-$.

Proof

We show $(s_n)$ converges. The subsequence $(s_{2n})$ is increasing, because $s_{2n+2} - s_{2n} = -a_{2n+2} + a_{2n+1} \ge 0$. Similarly $(s_{2n-1})$ is decreasing, since $s_{2n+1} - s_{2n-1} = a_{2n+1} - a_{2n} \le 0$. We claim

$$s_{2m} \le s_{2n+1} \qquad\text{for all } m, n \in \mathbb{N}. \tag{1}$$

First, $s_{2n} \le s_{2n+1}$ for all $n$, because $s_{2n+1} - s_{2n} = a_{2n+1} \ge 0$. If $m \le n$, then (1) holds because $s_{2m} \le s_{2n} \le s_{2n+1}$. If $m \ge n$, then (1) holds because $s_{2n+1} \ge s_{2m+1} \ge s_{2m}$.

By (1), $(s_{2n})$ is an increasing subsequence bounded above by every odd partial sum, and $(s_{2n+1})$ is a decreasing subsequence bounded below by every even partial sum. By Theorem 10.2 these converge, say to $s$ and $t$. Now

$$t - s = \lim_{n\to\infty} s_{2n+1} - \lim_{n\to\infty} s_{2n} = \lim_{n\to\infty} (s_{2n+1} - s_{2n}) = \lim_{n\to\infty} a_{2n+1} = 0,$$

so $s = t$, and it follows that $\lim_n s_n = s$.

For the last claim, note $s_{2k} \le s \le s_{2k+1}$, so both $s_{2k+1} - s$ and $s - s_{2k}$ are bounded by $s_{2k+1} - s_{2k} = a_{2k+1} \le a_{2k}$. Whether $n$ is even or odd, $|s - s_n| \le a_n$.

∎

Intuition

Each new term overshoots the limit in the opposite direction, by less than the previous overshoot. So the limit is trapped between any two consecutive partial sums, and the error is at most the first omitted term. For example, to compute $1 - \frac12 + \frac13 - \cdots$ to within $0.001$, it is enough to stop at a term $a_n \le 0.001$.

Common pitfallterms tending to 0 is not enough without “decreasing”

Alternating signs and $a_n \to 0$ do not suffice. Consider

$$1 - 1 + \tfrac12 - \tfrac14 + \tfrac13 - \tfrac19 + \tfrac14 - \tfrac1{16} + \cdots,$$

whose terms alternate in sign and tend to $0$ but do not decrease. Grouped in pairs, the partial sums are $\sum_{j=1}^k \left(\frac1j - \frac1{j^2}\right)$, and these tend to $+\infty$ because $\sum\frac1j$ diverges while $\sum\frac1{j^2}$ converges.

Common pitfallorder matters for conditionally convergent series

A series that converges but not absolutely is called conditionally convergent, and for such series the order of the terms matters. Take the terms of $1 - \frac12 + \frac13 - \frac14 + \cdots$ and write them as one positive term followed by two negative ones:

$$\begin{aligned} &\left(1 - \tfrac12\right) - \tfrac14 + \left(\tfrac13 - \tfrac16\right) - \tfrac18 + \left(\tfrac15 - \tfrac1{10}\right) - \tfrac1{12} + \cdots \\ &\quad = \tfrac12 - \tfrac14 + \tfrac16 - \tfrac18 + \tfrac1{10} - \tfrac1{12} + \cdots. \end{aligned}$$

Each group in parentheses is half of its first term, and the result is exactly half of the original series, so it converges to $\frac12\log 2$ instead of $\log 2$. The same terms in a different order give a different sum. Absolutely convergent series do not have this problem: every rearrangement has the same sum.

Try it: reorder the terms

Choose an order and watch where the partial sums go. For the conditionally convergent series you can even steer the sum to any target you like (Riemann’s rearrangement theorem). For the absolutely convergent series nothing you do changes the sum.

partial sumsoriginal sum 0.6931this order → 0.3466

Terms used600

Order: one positive term, then two negative terms, repeating. Terms: +1/1 −1/2 −1/4 +1/3 −1/6 −1/8 +1/5 −1/10 −1/12 +1/7 −1/14 −1/16 +1/9 −1/18 ⋯
After 600 terms the partial sum is 0.3459. The same terms, reordered, now sum to 0.3466, not 0.6931.

The terms of $1 - \frac12 + \frac13 - \frac14 + \cdots$ alternate in sign, and the absolute values $1/n$ add up to infinity. Does this series converge?

Yes, but only conditionallyNo, because the absolute values diverge

Try it: the zigzag

The partial sums of $1 - \frac12 + \frac13 - \cdots$ zigzag: even sums climb, odd sums fall, and every even sum sits below every odd sum, exactly as in (1). They squeeze down to $\log 2 \approx 0.6931$.

Terms n40

terms aₙpartial sums sₙ = a₁ + ⋯ + aₙsum = log 2

s_40 = 0.680803, which differs from the sum log 2 ≈ 0.693147 by 1.23e-2.

Converges by the alternating series theorem 15.3, but not absolutely. The partial sums zig-zag, and the limit lies between any two consecutive ones.

Reading alongside Abbott

Abbott proves the Alternating Series Test in Section 2.7, along with the theorem that rearranging an absolutely convergent series does not change its sum. In fact Abbott opens the whole chapter on sequences and series (Section 2.1) with the puzzle of rearrangements, as motivation for taking convergence seriously. Abbott does not use integral tests at this stage, since integrals come much later in that book. It settles the $p$-series with the Cauchy condensation test instead (see the note at the end of §14).

Check your understanding

Take the quiz (3 questions)

Which series converges?

$\sum \frac{1}{n^{1.01}}$$\sum \frac{1}{\sqrt n}$$\sum \frac{1}{n}$

How does $\sum \frac{(-1)^{n+1}}{n} = 1 - \frac12 + \frac13 - \cdots$ behave?

It converges, but not absolutely.It converges absolutely.It diverges.

For $\sum \frac{(-1)^{n+1}}{n}$, when does the bound in Theorem 15.3 guarantee $|s - s_n| \le 0.01$?

Once $n \ge 100$Once $n \ge 10$Once $n \ge 1000$

Exercises

Exercise 15.1

Determine which of the following series converge. Justify your answers.

1. $\sum \dfrac{(-1)^n}{n}$
2. $\sum \dfrac{(-1)^n n!}{2^n}$

Solution

(a) converges by the Alternating Series Theorem. (b) diverges: $\frac{n!}{2^n} \to \infty$, so the terms do not tend to $0$.

Tags: Warm-upSeries tests

Used in Exercise 15.2

Exercise 15.2

Repeat Exercise 15.1 for the following.

1. $\sum \left[\sin\left(\tfrac{n\pi}6\right)\right]^n$
2. $\sum \left[\sin\left(\tfrac{n\pi}7\right)\right]^n$

Solution

(a) diverges: for $n = 12j + 3$, $\sin(n\pi/6) = 1$, so the terms do not tend to $0$. (b) converges absolutely: $|\sin(n\pi/7)| \le \sin(3\pi/7) < 1$ for every $n$ (it is $0$ when $7 \mid n$), so compare with a geometric series.

Tags: StandardSeries tests

Exercise 15.3

Show $\displaystyle\sum_{n=2}^\infty \frac1{n(\log n)^p}$ converges if and only if $p > 1$.

Hint

Integral test with $f(x) = \frac1{x(\log x)^p}$; substitute $u = \log x$.

Solution

Let $f(x) = \dfrac1{x(\log x)^p}$ on $[2,\infty)$. For $p>0$, it is positive and decreasing, because $x(\log x)^p$ is positive and increasing. Substitute $u = \log x$: $\int_2^n f(x)\,dx = \int_{\log 2}^{\log n}u^{-p}\,du = \begin{cases}\log\log n - \log\log 2, & p = 1,\\[2pt] \dfrac{(\log n)^{1-p} - (\log 2)^{1-p}}{1-p}, & p \ne 1.\end{cases}$ As $n \to \infty$ this tends to $+\infty$ when $p = 1$ or $0<p<1$, and tends to the finite limit $\frac{(\log 2)^{1-p}}{p-1}$ when $p > 1$. By the integral test (Remark 15.2, Exercise 15.8) the series converges exactly when $p > 1$. For $p \le 0$, $\frac1{n(\log n)^p} \ge \frac1n$ for $n \ge 3$, so the series diverges by comparison with $\sum\frac1n$.

Tags: StandardSeries tests

Exercise 15.4

Determine which of the following series converge. Justify your answers.

1. $\sum_{n=2}^\infty \dfrac1{\sqrt n \log n}$
2. $\sum_{n=2}^\infty \dfrac{\log n}{n}$
3. $\sum_{n=4}^\infty \dfrac1{n (\log n)(\log \log n)}$
4. $\sum_{n=2}^\infty \dfrac{\log n}{n^2}$

Solution

(a) diverges ($\sqrt n \log n \le n$ for large $n$, so the terms dominate $\frac1n$). (b) diverges (terms $\ge \frac1n$ for $n \ge 3$). (c) diverges (integral is $\log\log\log x$). (d) converges ($\log n \le \sqrt n$, so terms $\le n^{-3/2}$).

Tags: StandardSeries tests

Exercise 15.5

Why didn’t we use the Comparison Test to prove Theorem 15.1 for $p > 1$?

Solution

To compare we need a known convergent series to compare against, and at that point we had only geometric series, which decay too fast to dominate $\frac1{n^p}$.

Tags: Warm-upExplain why

Exercise 15.6

1. Give an example of a divergent series $\sum a_n$ for which $\sum a_n^2$ converges.
2. Observe that if $\sum a_n$ is a convergent series of nonnegative terms, then $\sum a_n^2$ also converges. See Exercise 14.7.
3. Give an example of a convergent series $\sum a_n$ for which $\sum a_n^2$ diverges.

Solution

(a) $a_n = \frac1n$. (c) $a_n = \frac{(-1)^n}{\sqrt n}$.

Tags: StandardSeries testsExamples & counterexamples

Exercise 15.7

1. Prove that if $(a_n)$ is a decreasing sequence of real numbers and $\sum a_n$ converges, then $\lim n a_n = 0$.
2. Use (a) to give another proof that $\sum \frac1n$ diverges.

Hint

For (a), consider $|a_{N+1} + a_{N+2} + \cdots + a_n|$ for suitable $N$. Each term is at least $a_n$, so this is at least $(n - N)a_n$. Note the $a_n$ must be nonnegative, since otherwise the terms would not tend to $0$.

Solution

1. Since $\sum a_n$ converges, $a_n \to 0$; a decreasing sequence tending to $0$ has all terms $\ge 0$. Let $\varepsilon > 0$ and use the Cauchy criterion to choose $N$ with $\sum_{k=N+1}^n a_k < \varepsilon/2$ for all $n > N$. Because $(a_k)$ is decreasing, $a_k \ge a_n$ for $N < k \le n$, so $(n - N)a_n \le \sum_{k=N+1}^n a_k < \varepsilon/2$. For $n \ge 2N$ we have $n - N \ge n/2$, hence $\frac n2 a_n < \frac\varepsilon2$, i.e. $0 \le na_n < \varepsilon$. So $na_n \to 0$.
2. The sequence $a_n = \frac1n$ is decreasing but $na_n = 1 \not\to 0$. By (a), $\sum\frac1n$ cannot converge.

Tags: ChallengeSeries testsMonotone & Cauchy

Exercise 15.8

Formulate and prove a general integral test as advised in 15.2.

Hint

If $f \ge 0$ is decreasing on $[1,\infty)$ and $a_n = f(n)$, then $\int_1^{n+1} f \le \sum_{k=1}^n a_k \le a_1 + \int_1^n f$.

Solution

Integral test. Let $f \ge 0$ be a decreasing function on $[1, \infty)$ and let $a_n = f(n)$. Then $\sum a_n$ converges if and only if $\lim_{n\to\infty}\int_1^n f(x)\,dx < +\infty$; otherwise both the series and the integral limit are $+\infty$.

Proof. A decreasing function is integrable on each $[1,n]$. For $x \in [k, k+1]$ we have $a_{k+1} = f(k+1) \le f(x) \le f(k) = a_k$, so $a_{k+1} \le \int_k^{k+1}f \le a_k$. Summing over $k$, with $s_n = \sum_{k=1}^n a_k$ and $F(n) = \int_1^n f$, $s_n - a_1 \le F(n) \le s_{n-1}.$ Both $(s_n)$ and $(F(n))$ are increasing. If $(s_n)$ is bounded, so is $(F(n))$ (as $F(n) \le s_{n-1}$), so $F(n)$ converges. If $(F(n))$ is bounded, then $s_n \le a_1 + F(n)$ is bounded, so $\sum a_n$ converges. Hence one converges exactly when the other does, and otherwise both tend to $+\infty$. $\blacksquare$

Tags: ChallengeSeries tests

Used in Remark 15.2, Exercise 15.3