§14 Series

Our thorough treatment of sequences lets us obtain the basic properties of infinite series quickly.

Why this matters

An infinite sum is not something you can add up directly. Definition 14.2 turns it into a limit of ordinary finite sums, so every tool from §§7–12 applies at once. Series are also how many functions are actually computed: $e^x$, $\sin x$ and $\frac{1}{1-x}$ are all sums of power series (§23).

14.1 Summation Notation.

$\sum_{k=m}^n a_k$ is shorthand for $a_m + a_{m+1} + \cdots + a_n$. For example,

$$\sum_{k=2}^5 \frac1{k^2 + k} = \frac16 + \frac1{12} + \frac1{20} + \frac1{30}, \qquad \sum_{k=0}^n 2^{-k} = 1 + \frac12 + \frac14 + \cdots + \frac1{2^n}.$$

$\sum_{n=m}^\infty a_n$ is shorthand for $a_m + a_{m+1} + a_{m+2} + \cdots$, which we now give a meaning.

14.2 Infinite Series.

The partial sums of $\sum_{n=m}^\infty a_n$ are $s_n = a_m + a_{m+1} + \cdots + a_n = \sum_{k=m}^n a_k$. The series converges if $(s_n)$ converges to a real number $S$, and then we define $\sum_{n=m}^\infty a_n = S$. Thus

$$\sum_{n=m}^\infty a_n = S \quad\text{means}\quad \lim s_n = S, \quad\text{i.e.}\quad \lim_{n\to\infty} \sum_{k=m}^n a_k = S.$$

A series that does not converge diverges. We write $\sum a_n = +\infty$ if $\lim s_n = +\infty$, and similarly for $-\infty$. The symbol $\sum_{n=m}^\infty a_n$ has no meaning otherwise. When the starting index does not matter we just write $\sum a_n$.

If all $a_n \ge 0$, the partial sums increase, so by Theorems 10.2 and 10.4, $\sum a_n$ either converges or diverges to $+\infty$. In particular $\sum |a_n|$ is always meaningful. $\sum a_n$ converges absolutely if $\sum |a_n|$ converges. Absolutely convergent series converge (Corollary 14.7).

Intuition

Keep two sequences apart: the terms $(a_n)$ and the partial sums $(s_n)$. Convergence of a series is a statement about $(s_n)$. Whether a series converges depends only on its tail: dropping or changing finitely many terms changes every later partial sum by the same fixed amount. The value of the sum, of course, depends on every term.

Try it: watch the partial sums

Orange dots are the terms $a_n$ and blue dots the partial sums $s_n$. Compare $\sum 1/n$, whose terms go to $0$ but whose sums creep off to infinity, with $\sum 1/n^2$.

Terms n40

terms aₙpartial sums sₙ = a₁ + ⋯ + aₙ

s_40 = 4.278543.

Diverges, although very slowly: the partial sums grow like log n (Example 1 of §15, integral test). The terms tend to 0, so that alone does not guarantee convergence.

Example 1.

A series $\sum_{n=0}^\infty ar^n$ is a geometric series, the easiest kind to sum. For $r \ne 1$,

$$\sum_{k=0}^n ar^k = a\,\frac{1 - r^{n+1}}{1 - r}. \tag{1}$$

To verify this, multiply the left side by $1 - r$: $(a + ar + \cdots + ar^n) - (ar + ar^2 + \cdots + ar^{n+1}) = a - ar^{n+1}$. For $|r| < 1$, $r^{n+1} \to 0$ (Theorem 9.7(b)), so

$$\sum_{n=0}^\infty ar^n = \frac{a}{1 - r} \qquad\text{if } |r| < 1. \tag{2}$$

If $a \ne 0$ and $|r| \ge 1$, then $(ar^n)$ does not tend to $0$, so the series diverges by Corollary 14.5.

Example 2.

Formula (2) and the next fact are very important and should be used whenever possible, even though (1) is not proved until §15. For fixed $p > 0$,

$$\sum_{n=1}^\infty \frac1{n^p} \text{ converges if and only if } p > 1. \tag{1}$$

So for $p \le 1$ we can write $\sum 1/n^p = +\infty$. Exact values for $p > 1$ are hard. Fourier series or complex variables give

$$\sum_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}{6} = 1.6449\cdots, \tag{2}$$$$\sum_{n=1}^\infty \frac1{n^4} = \frac{\pi^4}{90} = 1.0823\cdots. \tag{3}$$

Similar formulas hold for every even $p$, but none is known for odd $p$. Nobody knows a formula for $\sum 1/n^3$, although it converges and can be approximated as closely as desired. It is often far easier to prove that a series converges than to find its value.

The Cauchy criterion

14.3 Definition.

A series $\sum a_n$ satisfies the Cauchy criterion if its partial sums form a Cauchy sequence:

$$\text{for each } \varepsilon > 0 \text{ there is } N \text{ such that } m, n > N \text{ implies } |s_n - s_m| < \varepsilon. \tag{1}$$

We lose nothing by requiring $n > m$, and writing $m - 1$ in place of $m$, (1) is equivalent to

$$\text{for each } \varepsilon > 0 \text{ there is } N \text{ such that } n \ge m > N \text{ implies } |s_n - s_{m-1}| < \varepsilon. \tag{2}$$

Since $s_n - s_{m-1} = \sum_{k=m}^n a_k$, this is

$$\text{for each } \varepsilon > 0 \text{ there is } N \text{ such that } n \ge m > N \text{ implies } \left| \sum_{k=m}^n a_k \right| < \varepsilon. \tag{3}$$

We usually use version (3).

Theorem 10.11 gives:

14.4 Theorem.

A series converges if and only if it satisfies the Cauchy criterion.

Used in Exercise 14.6, Exercise 14.9

14.5 Corollary.

If a series $\sum a_n$ converges, then $\lim a_n = 0$.

Used in Theorem 14.9, Exercise 14.7

Proof

Take $n = m$ in 14.3(3): for each $\varepsilon > 0$ there is $N$ such that $n > N$ implies $|a_n| < \varepsilon$.

∎

The converse fails: $\sum 1/n = +\infty$.

Common pitfallterms tending to 0 proves nothing

Corollary 14.5 is a test for divergence only. If $a_n \not\to 0$, the series diverges. If $a_n \to 0$, you have learned nothing: $\sum \frac1n$ diverges and $\sum \frac1{n^2}$ converges, and both have terms tending to $0$.

Convergence tests

14.6 Comparison Test.

Let $\sum a_n$ be a series with $a_n \ge 0$ for all $n$.

1. If $\sum a_n$ converges and $|b_n| \le a_n$ for all $n$, then $\sum b_n$ converges.
2. If $\sum a_n = +\infty$ and $b_n \ge a_n$ for all $n$, then $\sum b_n = +\infty$.

Proof

(i) For $n \ge m$, by the triangle inequality (Exercise 3.6(b)),

$$\left| \sum_{k=m}^n b_k \right| \le \sum_{k=m}^n |b_k| \le \sum_{k=m}^n a_k. \tag{1}$$

$\sum a_n$ satisfies the Cauchy criterion, so by (1) so does $\sum b_n$, which therefore converges.

(ii) If $(s_n)$ and $(t_n)$ are the partial sums of $\sum a_n$ and $\sum b_n$, then $t_n \ge s_n$, and $\lim s_n = +\infty$ forces $\lim t_n = +\infty$.

∎

Common pitfallcompare in the right direction

To prove convergence, bound your terms above by a convergent series. To prove divergence, bound them below by a divergent one. The other two combinations prove nothing: knowing $\frac1{n^2} \le \frac1n$, where $\sum\frac1n$ diverges, says nothing about $\sum\frac1{n^2}$.

14.7 Corollary.

Absolutely convergent series are convergent.

Used in Exercise 14.6

Proof

If $\sum |b_n|$ converges, apply 14.6(i) with $a_n = |b_n|$.

∎

The proofs of this corollary and of the Alternating Series Theorem 15.3 use the completeness of $\mathbb{R}$.

The Ratio Test is popular because it is often easy to use. But it is less general than the Root Test, the Root Test is what determines the radius of convergence of a power series, and the Ratio Test is useless if some $a_n = 0$. To review $\limsup$ and $\liminf$, see Definition 10.6, Theorems 10.7 and 11.8, and §12.

For $\sum 1/n$ and for $\sum 1/n^2$, the ratio $|a_{n+1}/a_n|$ tends to $1$ in both cases. One series diverges and the other converges. What can a ratio test say when that limit is $1$?

It says the series convergesIt says the series divergesIt cannot decide: a limit of $1$ is consistent with both outcomes

14.8 Ratio Test.

A series $\sum a_n$ of nonzero terms

1. converges absolutely if $\limsup |a_{n+1}/a_n| < 1$,
2. diverges if $\liminf |a_{n+1}/a_n| > 1$.
3. Otherwise $\liminf |a_{n+1}/a_n| \le 1 \le \limsup |a_{n+1}/a_n|$ and the test gives no information.

If $\lim |a_{n+1}/a_n|$ exists, it equals both the lim sup and the lim inf, so the Ratio Test decides the question unless that limit is $1$. The proof comes after the Root Test.

14.9 Root Test.

Let $\sum a_n$ be a series and $\alpha = \limsup |a_n|^{1/n}$. The series $\sum a_n$

1. converges absolutely if $\alpha < 1$,
2. diverges if $\alpha > 1$.
3. Otherwise $\alpha = 1$ and the test gives no information.

Proof

(i) Choose $\varepsilon > 0$ with $\alpha + \varepsilon < 1$. By Definition 10.6 there is $N$ with $\alpha - \varepsilon < \sup\{|a_n|^{1/n} : n > N\} < \alpha + \varepsilon$, so $|a_n| < (\alpha + \varepsilon)^n$ for $n > N$. The geometric series $\sum_{n > N} (\alpha + \varepsilon)^n$ converges, so $\sum_{n > N} a_n$ converges absolutely by comparison, and so does $\sum a_n$ (Exercise 14.9).

(ii) If $\alpha > 1$, then by Theorem 11.7 a subsequence of $(|a_n|^{1/n})$ tends to $\alpha > 1$, so $|a_n| > 1$ for infinitely many $n$. Then $a_n \not\to 0$, and the series diverges by Corollary 14.5.

(iii) For both $\sum \frac1n$ and $\sum \frac1{n^2}$, $\alpha = 1$ by Theorem 9.7(c). One diverges and the other converges, so $\alpha = 1$ decides nothing.

∎

Proof of the Ratio Test

Let $\alpha = \limsup |a_n|^{1/n}$. By Theorem 12.2,

$$\liminf \left|\frac{a_{n+1}}{a_n}\right| \le \alpha \le \limsup \left|\frac{a_{n+1}}{a_n}\right|. \tag{1}$$

If $\limsup |a_{n+1}/a_n| < 1$ then $\alpha < 1$, and the Root Test gives convergence. If $\liminf |a_{n+1}/a_n| > 1$ then $\alpha > 1$, and the Root Test gives divergence. Part (iii) again follows from $\sum \frac1n$ and $\sum \frac1{n^2}$.

∎

Inequality (1) shows the Root Test is superior: whenever the Root Test gives no information ($\alpha = 1$), neither does the Ratio Test. Example 8 below is a series the Ratio Test cannot handle but the Root Test can. Usually, though, the tests fail together:

Intuition

Both tests are comparisons with a geometric series. If $|a_n|^{1/n}$ eventually stays below some $r < 1$, then $|a_n| < r^n$, and $\sum r^n$ converges. The Ratio Test reaches the same comparison one step at a time.

Common pitfallratios below 1 are not enough

“$\left|\frac{a_{n+1}}{a_n}\right| < 1$ for every $n$” does not imply convergence. For $\sum\frac1n$ the ratios $\frac{n}{n+1}$ are all less than $1$, yet the series diverges. The ratios must stay below some fixed $r < 1$, which is what $\limsup\left|\frac{a_{n+1}}{a_n}\right| < 1$ guarantees.

14.10 Remark.

If the $a_n$ are nonzero and $\lim |a_{n+1}/a_n| = 1$, then $\alpha = \limsup |a_n|^{1/n} = 1$ by Corollary 12.3, so neither the Ratio Test nor the Root Test gives information about $\sum a_n$.

There is no clear-cut strategy for choosing among Comparison, Ratio and Root (and the two tests of §15). If nothing in the form of $\sum a_n$ suggests a test and the ratios $a_{n+1}/a_n$ are easy to compute, you may as well try the Ratio Test first.

Example 3.

$\displaystyle\sum_{n=2}^\infty \left(-\frac13\right)^n = \frac19 - \frac1{27} + \frac1{81} - \frac1{243} + \cdots$ is geometric: it is $\frac19 \sum_{n=0}^\infty \left(-\frac13\right)^n$, so its sum is $\dfrac{1/9}{1 - (-1/3)} = \dfrac1{12}$. It also converges by comparison with $\sum (1/3)^n$, which converges by the Ratio or Root Test (both limits are $\frac13$). But none of these tests gives the exact value.

Example 4.

$\displaystyle\sum \frac{n}{n^2 + 3}$. With $a_n = \frac{n}{n^2+3}$, $\frac{a_{n+1}}{a_n} = \frac{n+1}{n} \cdot \frac{n^2 + 3}{n^2 + 2n + 4} \to 1$, so Ratio and Root Tests give no information (Remark 14.10). Since $a_n \approx \frac1n$ we expect divergence, and indeed $\frac{n}{n^2+3} \ge \frac{n}{4n^2} = \frac1{4n}$. Since $\sum \frac1{4n}$ diverges (its partial sums are a quarter of those of $\sum \frac1n$), the series diverges by comparison.

Example 5.

$\displaystyle\sum \frac1{n^2 + 1}$. Ratio and Root Tests give no information, but $\frac1{n^2+1} \le \frac1{n^2}$ and $\sum \frac1{n^2}$ converges, so the series converges by comparison.

Example 6.

$\displaystyle\sum \frac{n}{3^n}$. Here $\frac{a_{n+1}}{a_n} = \frac{n+1}{3n} \to \frac13$, so the series converges by the Ratio Test. The Root Test works just as easily using $\lim n^{1/n} = 1$.

Example 7.

$\displaystyle\sum a_n$ with $a_n = \left[\frac{2}{(-1)^n - 3}\right]^n$. The form suggests the Root Test: $|a_n|^{1/n} = 1$ for even $n$ and $\frac12$ for odd $n$, so $\alpha = 1$ and the test fails, as does the Ratio Test. But an alert reader notices $a_n = 1$ for even $n$, so $a_n \not\to 0$ and the series diverges by Corollary 14.5.

Example 8.

$\displaystyle\sum_{n=0}^\infty 2^{(-1)^n - n} = 2 + \frac14 + \frac12 + \frac1{16} + \frac18 + \frac1{64} + \cdots$. Since $a_n \le \frac1{2^{n-1}}$, it converges by comparison. Its real interest: $\frac{a_{n+1}}{a_n} = \frac18$ for even $n$ and $2$ for odd $n$, so $\liminf = \frac18 < 1 < 2 = \limsup$ and the Ratio Test gives no information. But $a_n^{1/n} = 2^{\frac1n - 1}$ for even $n$ and $2^{-\frac1n - 1}$ for odd $n$, both tending to $\frac12$. So $\alpha = \frac12 < 1$ and the Root Test gives convergence.

Example 9.

$\displaystyle\sum_{n=1}^\infty \frac{(-1)^{n+1}}{\sqrt n} = 1 - \frac1{\sqrt2} + \frac1{\sqrt3} - \frac12 + \cdots$. Ratio and Root Tests fail. Since $\sum \frac1{\sqrt n}$ diverges, 14.6(i) cannot prove convergence, and since the terms change sign, 14.6(ii) cannot prove divergence. The series in fact converges by the Alternating Series Test 15.3.

Reading alongside Abbott

Abbott introduces series in Section 2.4, where the Monotone Convergence Theorem settles $\sum\frac1{n^2}$ and the harmonic series. There Abbott proves the Cauchy condensation test: for decreasing nonnegative terms, $\sum a_n$ converges exactly when $\sum 2^k a_{2^k}$ does. With $a_n = \frac1{n^p}$ the condensed series is geometric with ratio $2^{1-p}$, which gives the result of §15 with no integrals at all. Section 2.7 has the Cauchy criterion, comparison, absolute convergence and alternating series. The Ratio Test appears there as an exercise, and the exact value $\frac{\pi^2}{6}$ of Example 2 is derived in Section 8.3.

Try it: which test works?

Pick a series and step through the four tests. Notice that no single test settles everything: the Ratio Test fails on the $p$-series and on the oscillating series, the Root Test still works on the latter, and only the Alternating Series Test (Theorem 15.3) handles $\sum(-1)^{n+1}/n$.

|a(n+1)/a(n)|level 1 (dashed)

Terms plotted30

← PreviousRatio TestRoot TestComparison TestAlternating Series TestNext →

Ratio Test: Inconclusive. lim |a_{n+1}/a_n| = 1. The limit is 1, so 14.8(iii) says the test gives no information.

All four tests on Σ (−1)ⁿ⁺¹ / n

Test	Result
Ratio Test	Inconclusive
Root Test	Inconclusive
Comparison Test	Inconclusive
Alternating Series Test	Conclusive: converges

Practice: fill in the gaps

The proof of part (i) of the Ratio Test (Theorem 14.8): trap the terms under a geometric series.

Fill in the gaps: Ratio Test, part (i) (Theorem 14.8)

This is the skeleton of the proof. In each gap choose the option that makes the argument work, then check.

1. Let $L = \limsup |a_{n+1}/a_n| < 1$ and pick a number $r$ with $L < r < 1$$r > 1$$r < L$.
2. Since $r > L$, there is $N$ such that $|a_{n+1}/a_n| < r$ for all $n \ge N$.
3. Multiplying these inequalities, for $n \ge N$ we get $|a_n| \le$ $|a_N|\, r^{\,n-N}$$|a_N|\, / \, r^{\,n-N}$$|a_N|\, n\, r$.
4. The geometric series $\sum r^n$ converges because $0 < r < 1$. By the Comparison Test, Theorem 14.6(i)the Alternating Series Theorem 15.3the Integral Test, $\sum |a_n|$ converges, so $\sum a_n$ converges absolutely.

Practice: spot the error

Two arguments, each with one line that claims more than the test allows.

Spot the error: the harmonic series

One line of this argument is wrong. Press the button on the line you think is the mistake.

1. Let $a_n = \frac1n$. For every $n$, $\left|\frac{a_{n+1}}{a_n}\right| = \frac{n}{n+1} < 1$.
2. By the Ratio Test, a series whose ratios are always less than $1$ converges.
3. So $\sum \frac1n$ converges.

Spot the error: comparing with the harmonic series

One line of this argument is wrong. Press the button on the line you think is the mistake.

1. For every $n \ge 1$, $\frac{1}{n^2 + n} \le \frac1n$.
2. The series $\sum \frac1n$ diverges.
3. By the Comparison Test, $\sum \frac{1}{n^2 + n}$ diverges.

Check your understanding

Take the quiz (3 questions)

If $\lim a_n = 0$, must $\sum a_n$ converge?

Yes, by Corollary 14.5.No: $\sum \frac1n$ diverges although $\frac1n \to 0$.

What does the Ratio Test say about $\sum \frac{1}{n^2}$?

Nothing: the ratios tend to $1$.It converges, because the ratios are less than $1$.It diverges.

What is $\sum_{n=0}^\infty \left(\frac23\right)^n$?

$3$$2$$\frac23$

Exercises

Exercise 14.1

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

1. $\sum \dfrac{n^4}{2^n}$
2. $\sum \dfrac{2^n}{n!}$
3. $\sum \dfrac{n^2}{3^n}$
4. $\sum \dfrac{n!}{n^4 + 3}$
5. $\sum \dfrac{\cos^2 n}{n^2}$
6. $\sum_{n=2}^\infty \dfrac1{\log n}$

Solution

(a) converges (ratio $\to \frac12$) (b) converges (ratio $\to 0$) (c) converges (ratio $\to \frac13$) (d) diverges (terms $\to \infty$) (e) converges (compare with $\frac1{n^2}$) (f) diverges ($\frac1{\log n} > \frac1n$)

Tags: Warm-upSeries tests

Used in Exercise 14.2, Exercise 14.3, Exercise 14.4

Exercise 14.2

Repeat Exercise 14.1 for the following.

1. $\sum \dfrac{n-1}{n^2}$
2. $\sum (-1)^n$
3. $\sum \dfrac{3n}{n^3}$
4. $\sum \dfrac{n^3}{3^n}$
5. $\sum \dfrac{n^2}{n!}$
6. $\sum \dfrac1{n^n}$
7. $\sum \dfrac{n}{2^n}$

Solution

(a) diverges ($\frac{n-1}{n^2} \ge \frac1{2n}$ for $n \ge 2$) (b) diverges (terms don’t tend to $0$) (c) converges ($= 3\sum\frac1{n^2}$) (d)–(g) converge (ratio or root test)

Tags: Warm-upSeries tests

Exercise 14.3

Repeat Exercise 14.1 for the following.

1. $\sum \dfrac1{\sqrt{n!}}$
2. $\sum \dfrac{2 + \cos n}{3^n}$
3. $\sum \dfrac1{2^n + n}$
4. $\sum \left(\tfrac12\right)^n (50 + n^2)$
5. $\sum \sin\dfrac{n\pi}9$
6. $\sum \dfrac{(100)^n}{n!}$

Solution

(a) converges (ratio $\frac1{\sqrt{n+1}} \to 0$) (b) converges ($\le 3 \cdot 3^{-n}$) (c) converges ($\le 2^{-n}$) (d) converges (e) diverges (terms don’t tend to $0$) (f) converges (ratio $\frac{100}{n+1} \to 0$)

Tags: Warm-upSeries tests

Exercise 14.4

Repeat Exercise 14.1 for the following.

1. $\sum_{n=2}^\infty \dfrac1{[n + (-1)^n]^2}$
2. $\sum \left[\sqrt{n+1} - \sqrt n\right]$
3. $\sum \dfrac{n!}{n^n}$

Solution

(a) converges ($[n + (-1)^n]^2 \ge (n-1)^2$) (b) diverges (it telescopes: $s_n = \sqrt{n+1} - 1 \to \infty$) (c) converges (ratio $\to \frac1e$)

Tags: StandardSeries tests

Exercise 14.5

Suppose $\sum a_n = A$ and $\sum b_n = B$ with $A, B$ real. Use limit theorems from §9 to quickly prove:

1. $\sum(a_n + b_n) = A + B$.
2. $\sum ka_n = kA$ for $k \in \mathbb{R}$.
3. Is $\sum a_n b_n = AB$ a reasonable conjecture? Discuss.

Solution

(c) No. For $a_n = b_n = \frac1{2^n}$ ($n \ge 1$), $A = B = 1$ but $\sum a_n b_n = \sum \frac1{4^n} = \frac13$.

Tags: Warm-upSeries testsLimit laws

Used in Exercise 14.6, Exercise 14.8, Exercise 16.3

Exercise 14.6

1. Prove that if $\sum |a_n|$ converges and $(b_n)$ is bounded, then $\sum a_n b_n$ converges.
2. Observe that Corollary 14.7 is a special case of part (a).

Hint

Use Theorem 14.4, or compare $|a_n b_n| \le M|a_n|$.

Solution

1. Choose $M$ with $|b_n| \le M$ for all $n$. Then $|a_nb_n| \le M|a_n|$. The series $\sum M|a_n|$ converges (it is $M$ times a convergent series, Exercise 14.5(b)), so $\sum a_nb_n$ converges by the Comparison Test 14.6(i) (with $M|a_n|$ in the role of the dominating series).
2. Take $b_n = 1$ for all $n$: if $\sum|a_n|$ converges then $\sum a_n$ converges, which is Corollary 14.7.

Tags: StandardSeries tests

Exercise 14.7

Prove that if $\sum a_n$ is a convergent series of nonnegative numbers and $p > 1$, then $\sum a_n^p$ converges.

Hint

$a_n \to 0$, so $a_n \le 1$ eventually, and then $a_n^p \le a_n$.

Solution

Since $\sum a_n$ converges, $a_n \to 0$ (Corollary 14.5), so there is $N$ with $0 \le a_n \le 1$ for $n > N$. For such $n$, $a_n^p \le a_n$ because $p > 1$ and $0 \le a_n \le 1$. By the Comparison Test, $\sum_{n>N} a_n^p$ converges, and so does $\sum a_n^p$ (adding finitely many terms does not affect convergence, Exercise 14.9).

Tags: StandardSeries tests

Used in Exercise 15.6

Exercise 14.8

Show that if $\sum a_n$ and $\sum b_n$ are convergent series of nonnegative numbers, then $\sum \sqrt{a_n b_n}$ converges.

Hint

$\sqrt{a_n b_n} \le a_n + b_n$ for all $n$.

Solution

Since $(\sqrt{a_n} - \sqrt{b_n})^2 \ge 0$, we get $\sqrt{a_nb_n} \le \frac12(a_n + b_n) \le a_n + b_n$. The series $\sum(a_n + b_n)$ converges by Exercise 14.5(a), so $\sum\sqrt{a_nb_n}$ converges by the Comparison Test.

Tags: StandardSeries tests

Exercise 14.9

The convergence of a series does not depend on any finite number of its terms, though its value does. Precisely: if $\{n \in \mathbb{N} : a_n \ne b_n\}$ is finite, then $\sum a_n$ and $\sum b_n$ both converge or both diverge. Prove this.

Hint

This is almost obvious from Theorem 14.4: beyond the last index where they differ, the sums in 14.3(3) are identical.

Solution

Let $s_n$ and $t_n$ be the partial sums of $\sum a_n$ and $\sum b_n$, and choose $N_0$ larger than every index at which $a_n \ne b_n$. For $n \ge m > N_0$ we have $\sum_{k=m}^n a_k = \sum_{k=m}^n b_k$. Hence, for any $\varepsilon$, the Cauchy condition 14.3(3) for $\sum a_n$ holds with some $N \ge N_0$ if and only if it holds for $\sum b_n$ with the same $N$. By Theorem 14.4 the two series converge or diverge together. (When they converge, $s_n - t_n$ is constant for $n \ge N_0$, so the sums differ by that constant, which is why the value, unlike convergence, depends on the early terms.)

Tags: StandardSeries tests

Used in Theorem 14.9, Exercise 14.7

Exercise 14.10

Find a series $\sum a_n$ that diverges by the Root Test but for which the Ratio Test gives no information. Compare Example 8.

Solution

$a_n = 2^{n + (-1)^n}$: the ratios alternate between $\frac12$ and $8$, but $a_n^{1/n} \to 2$.

Tags: StandardSeries testsExamples & counterexamples

Exercise 14.11

Let $(a_n)$ be a sequence of nonzero reals such that the ratios $\left(\frac{a_{n+1}}{a_n}\right)$ form a constant sequence. Show $\sum a_n$ is a geometric series.

Solution

Let $r$ be the common value of $\frac{a_{n+1}}{a_n}$, so $a_{n+1} = ra_n$ for all $n$. By induction $a_n = a_1r^{n-1}$ (true for $n = 1$, and $a_{n+1} = ra_n = a_1r^n$). So $\sum a_n = \sum_{n=1}^\infty a_1r^{n-1}$, a geometric series with first term $a_1$ and ratio $r$.

Tags: Warm-upSeries tests

Exercise 14.12

Let $(a_n)_{n\in\mathbb{N}}$ be a sequence with $\liminf |a_n| = 0$. Prove there is a subsequence $(a_{n_k})$ such that $\sum_{k=1}^\infty a_{n_k}$ converges.

Hint

Choose $n_k$ increasing with $|a_{n_k}| < 2^{-k}$, which is possible by Theorem 11.7 or 11.2.

Solution

Since $\liminf|a_n| = 0$, Theorem 11.7 gives a subsequence $(a_{m_j})$ with $|a_{m_j}| \to 0$. Choose indices inductively: given $n_{k-1}$, pick $j$ with $m_j > n_{k-1}$ and $|a_{m_j}| < 2^{-k}$ (possible because $|a_{m_j}| \to 0$ and $m_j \to \infty$), and let $n_k = m_j$. Then $n_1 < n_2 < \cdots$ and $|a_{n_k}| < 2^{-k}$. Since $\sum 2^{-k}$ converges, the Comparison Test shows $\sum_k a_{n_k}$ converges.

Tags: ChallengeSeries testsSubsequences

Exercise 14.13

It is often much harder to find the value of an infinite sum than to show it exists. Here are some sums that can be handled.

1. Calculate $\sum_{n=1}^\infty \left(\frac23\right)^n$ and $\sum_{n=1}^\infty \left(-\frac23\right)^n$.
2. Prove $\sum_{n=1}^\infty \frac1{n(n+1)} = 1$. (Note $\sum_{k=1}^n \frac1{k(k+1)} = \sum_{k=1}^n \left[\frac1k - \frac1{k+1}\right]$.)
3. Prove $\sum_{n=1}^\infty \frac{n-1}{2^{n+1}} = \frac12$. (Note $\frac{k-1}{2^{k+1}} = \frac{k}{2^k} - \frac{k+1}{2^{k+1}}$.)
4. Use (c) to calculate $\sum_{n=1}^\infty \frac{n}{2^n}$.

Solution

(a) $2$ and $-\frac25$. (b) The partial sums telescope to $1 - \frac1{n+1}$. (c) They telescope to $\frac12 - \frac{n+1}{2^{n+1}} \to \frac12$. (d) $\frac{n}{2^n} = 2 \cdot \frac{n-1}{2^{n+1}} + \frac1{2^n}$, so the sum is $2 \cdot \frac12 + 1 = 2$.

Tags: Warm-upSeries testsComputation

Exercise 14.14

Prove $\sum_{n=1}^\infty \frac1n$ diverges by comparing with $\sum_{n=2}^\infty a_n$, where

$$(a_n) = \left(\tfrac12, \tfrac14, \tfrac14, \tfrac18, \tfrac18, \tfrac18, \tfrac18, \tfrac1{16}, \dots, \tfrac1{16}, \tfrac1{32}, \dots\right)$$

has $2^{k-1}$ copies of $\frac1{2^k}$.

Hint

$\frac1n \ge a_n$, and each block of $2^{k-1}$ terms $\frac1{2^k}$ adds up to $\frac12$.

Solution

The $k$th block of $(a_n)$ consists of the $2^{k-1}$ terms $\frac1{2^k}$, in positions $n = 2^{k-1}+1, \dots, 2^k$. For these $n$ we have $n \le 2^k$, so $\frac1n \ge \frac1{2^k} = a_n$. Each block has sum $2^{k-1}\cdot\frac1{2^k} = \frac12$, so $\sum_{n=2}^{2^K} a_n = \frac K2 \to +\infty \quad (K \to \infty).$ The partial sums of $\sum a_n$ are increasing and unbounded, so $\sum_{n \ge 2} a_n = +\infty$, and by the Comparison Test $\sum_{n \ge 2}\frac1n = +\infty$. Hence $\sum_{n \ge 1}\frac1n$ diverges.

Tags: StandardSeries tests

Used in §15