112
9. SEQUENCES AND SERIES
18. Prove that the sum of the alternating harmonic series is
∞
X
(−1)n−1
n=1
n
= ln 2
Hint: Show that a partial sum of the alternating harmonic series is
s2n = h2n − hn where hn = an + ln n and the sequence {an } is defined
in Exercise 16 in Section 1.4. Then use the result of the latter exercise
to prove that sn → ln 2 as n → ∞.
60. Ratio and root tests
60.1. Absolutely convergent series.
Definition
13. (Absolute convergence)
P
A seriesP an is called absolutely convergent if the series of absolute
values
|an | is convergent.
The absolute convergence is stronger than convergence, meaning
that there are convergent series that do not
For
Pconverge absolutely.
n−1
example, the alternating harmonic series
an , an = (−1) /n, is
convergent, but not absolutely convergent because the series
P of absolute
values |an | = 1/n is nothing but the harmonic series
1/n which is
divergent (as a p−series with p = 1). On the other hand, the absolute
convergence implies convergence.
Theorem 38. (Convergence and absolute convergence)
Every absolutely convergent series is convergent
Proof: For any sequence {an }, the following inequality holds
0 ≤ an + |an | ≤ 2|an |
P
because |an | is either an or −an. It shows that the series P
bn , where
bnP= an + |an |, converges
by
the
comparison
test
because
2|an | =
P
2P |an | P
converges
an converges absolutely. Hence, the series
P if
an =
bn − |an | converges as the difference of two convergent
series.
P
Example 91. Test the series [sin(n)−2 cos(2n)]/n3/2 for absolute
convergence.
Solution: Making use of the inequality |A + B| ≤ |A| + |B| and the
properties that | sin(x)| ≤ 1 and | cos(x)| ≤ 1, one infers
|an | =
| sin(n)| + 2| cos(2n)|
3
| sin(n) − 2 cos(2n)|
≤
≤ 3/2
3/2
3/2
n
n
n
60. RATIO AND ROOT TESTS
113
P
The series of absolute values
P −3/2 |an | converges by comparison with the
convergent p−series 3 n
(here p = 3/2 > 1). So the series in
question converges absolutely. 2
Definition
14. (Conditional convergence)
P
A series
an is called conditionally convergent if it is convergent
but not absolutely convergent.
Thus, all convergent series are separated into two classes of conditionally and absolutely convergent series. The key difference between
properties of absolutely and conditionally convergent series is studied
in the next section.
60.2. The ratio test.
Theorem 39.
P (The ratio test)
Given a series
an , suppose the following limit exists
a n+1 lim =c
n→∞ an
where c ≥ 0 or c = ∞. Then
P
• If c < 1, P an converges absolutely
• If c > 1,
an diverges
• If c = 1, the test gives no information
Proof: If c < 1, then the existence of the limit means that for any
ε > 0 there is an integer N such that
a a n+1 n+1 −
c
<
ε
=⇒
−ε < < c + ε = q < 1 for all n ≥ N
an
an
Note that since c is strictly less than 1, one can always take ε > 0
small enough so that the number q = c + ε < 1. In particular, put
n = N + k − 1, where k ≥ 2. Applying the inequality |an+1 | < q|an|
cosequtively k times,
|aN +k | < q|aN +k−1| < q 2|aN +k−2 | < · · · < q k |aN | = |aN |q −N q N +k
This shows that
|an | < βq n , β = |aN |q −N , for all n ≥ N
P
The series P
|an | converges
P by comparison with the
P convergent geometric series
βq n = β q n because q < 1. So,
an converges absolutely. If c > 1, then there is an integer N such that |an+1 |/|an | > 1
or |an+1 | > |an | ≥ 0 for all n ≥ N. Hence, the necessary condition for
aPseries to converge, an → 0 as n → ∞ does not hold, that is, the series
an diverges. If c = 1, it is sufficient to give examples of a convergent
(55)
114
9. SEQUENCES AND SERIES
P −p
and divergent series for which c = 1. Consider a p−series
n . One
has
a np
1
n+1 c = lim = lim
=1
= lim
n→∞ an
n→∞ (n + 1)p
n→∞ (1 + 1/n)p
for any p. But a p−series converges if p > 1 and diverges otherwise.
92. Find all values of p and q for which the series
P∞Example
p n
n=1 n q converges absolutely.
Solution: Here an = np q n . One has
a (n + 1)p |q|n+1
(1 + 1/n)p
n+1 c = lim =
|q|
lim
= |q|
= lim
n→∞ an
n→∞
n→∞
np
|q|n
1
So, for |q| < 1 and any p the series converges absolutely by the ratio
test. If q = ±1, the ratio test is inconclusiveP
and theseP
cases have
P to be
p
studied by different means. If |q| = 1, then
|an | =
n =
1/n−p
which is a p−series that converges if −p > 1 or p < −1. Thus, the
series converges absolutely for all p if |q| < 1 and for p < −1 if q = ±1.
Note that for −1 ≤ p < 0 and q = −1 the series conditionally converges
(i.e. it is convergent but not absolutely
P convergent). In this case, it
is a convergent alternating p−series (−1)n /n−p (see Exercise 14 in
Section 1.6). 2
60.3. The root test.
Theorem 40.
P (The root test)
Given a series
an , suppose the following limit exists
p
lim n |an | = c
n→∞
where c ≥ 0 or c = ∞. Then
P
• If c < 1, P an converges absolutely
• If c > 1,
an diverges
• If c = 1, the test gives no information
Proof: If c < 1, then, as in the proof of the ratio test, the existence
of the limit means that for any c < q < 1 there is an integer N such
p
n
|an | < q =⇒ |an | < q n for all n ≥ N
P
This shows that the series P|an | converges by comparison
with the
P
n
convergent geometric series
q , 0 < q < 1. So,
an p
converges
n
absolutely. If c > 1, then there exists an integer N such that |an | > 1
for all n ≥ N and, hence,
the condition an → 0 as n → ∞ does
P
not
hold. The series
an diverges. If c = 1, consider a p−series:
√
√
n
n−p = ( n n)−p → 1−p = 1 by Theorem 1.6. But a p−series converges
if p > 1 and diverges if p < 1. So the root test is inconclusive.
60. RATIO AND ROOT TESTS
Example 93. Test the convergence of the series
[(2n2 + 5)/(3n2 + 2)]n
115
P
an where an =
Solution: Here |an | = an and the absolute convergence is equivalent
to the convergence. One has
lim
n→∞
p
n
2 + 5/n2
2
2n2 + 5
|an | = lim
=
lim
=
<1
n→∞ 3 + 2/n2
n→∞ 3n2 + 2
3
So the series converges. 2
60.4. Oscillatory behavior of sequences in the root and
ratio tests. Consider a sequence defined recursively by P
a1 = 1 and
1
an+1 = 2 sin(n)an . An attempt to test the convergence of
an by the
ratio test leads to the sequence cn = |an+1 |/|an | = 21 | sin(n)| that does
not converge as it oscillates between 0 and 1/2. Similarly, the sequence
used in the root test may also have an oscillatory
behavior and be nonp
1
n
n
convergent, e.g., an = ( 2 sin(n)) so that cn = |an | = 21 | sin(n)|. The
ratio and root tests, as stated in Theorems and 40, assume the existence of the limit (cn → c). What can be said about the convergence
of a series when this limit does not exist?
To answer this question, recall that in the proof of the ratio or root
test the existence of limn→∞ cn = c < 1 has been used only to establish
the boundedness
that cn < 1 for all n ≥ N which is sufficient for
P
the series
an to converge. But the boundedness property does not
imply the convergence! Evidently, the boundedness condition holds
in the above examples, cn = 21 | sin(n)| ≤ 12 < 1 for all n. Similarly,
the existence
of the limit value c > 1 has only been used to show
p
n
that
|an | ≥ 1 or |an| ≥ 1 for infinitely many n P
to conclude that
the sequence {an } cannot converge to 0 and, hence,
an diverges. If
|an+1 |/|an | ≥ 1 for all n ≥ N, then again {an } cannot converge to 0
(by the proof of the ratio test). Thus, the convergence of {cn } in the
root or ratio test is not really necessary.
Theorem 41. (Ratio and root tests refined)
p
P
Given a series n an , put cn = |an+1 |/|an | or cn = n |an |. Then
X
cn < 1 for all n ≥ N
=⇒
an converges
( p
n
X
|an | ≥ 1 for infinitely many n
=⇒
an diverges
|an+1 |
≥ 1 for all n ≥ N
|an |
for some integer N.
116
9. SEQUENCES AND SERIES
60.5. Wider scope of the p
root test. If the limit of |an+1 |/|an |
n
exists, then so does the limit of |an | and
p
|an+1 |
(56)
lim n |an | = lim
n→∞
n→∞ |an |
p
The converse is not true, that is, the existence of the limit of n |an |
does not generally imply the existence of the limit of |an+1p
|/|an | (the
latter may or may not exist). Furthermore, if the sequence n |an | does
not converge, neither does |an+1 |/|an |. A proof of these assertions is
given in more advanced calculus courses. Thus, the ratio test have the
same predicting power as the root test only if |an+1 |/|an | converges.
In general, the root test (as in Theorem 41) has wider scope, meaning that whenever the ratio test shows convergence, the root test does
too, and whenever the root test is inconclusive, the ratio test is too.
The subtlety to note here is that the converse of the latter statement
is not generally true, that is, the inconclusiveness of the ratio test does
not imply the inconclusiveness of the root test. The assertion can be
illustrated with the following example. Consider a convergent series
obtained from the sum of two geometric series in which the order of
summation is changed:
∞
∞
∞
X
1 1
1
1
1
1
1X 1
1X 1
3
an = + + 2 + 2 + 3 + 3 + · · · =
+
=
2 3 2
3
2
3
2 k=0 2k 3 n=0 3k
2
n=1
where the sum of a geometric series has been used (Theorem 27). Now
note that if n = 2k is even, then a2k = (1/2)k and a2k−1 = (1/3)k if
n = 2k − 1 is odd. Take the subsequence of ratios for even n = 2k,
c2k = a2k+1 /a2k = (2/3)k /9. It converges to 0 as k → ∞. On the
other hand, the subsequence of ratios for odd n = 2k − 1 diverges:
c2k−1 = (3/2)k → ∞ as k → ∞. So, the limit of cn does not exist and,
moreover, the ratio test (as in Theorem 41) fails miserably because cn is
not even bounded.
The series converges by the root
√
√ test. Indeed, c2k =
√
√
2k a
2k−1 a
=
1/
2
<
1
and
c
=
=
1/
3 < 1. Although the
2k
2k−1
2k−1
√
√
sequence cn does not converge (it oscillates between 1/ 3 and 1/ 2), it
is bounded cn < 1 for all n and, hence, the series converges by Theorem
41. A similar example is given in Exercise 19. Thus, the ratio test is
sensitive to the order of summation, while this is not so for the root
test.
60.6. When the ratio test is inconclusive.
Theorem
42. (De Morgan’s ratio test)
P
Let
an be a series in which |an+1 |/|an | → 1 as n → ∞. The series
60. RATIO AND ROOT TESTS
117
converges absolutely if
a n+1 lim n − 1 = b < −1
n→∞
an
A proof of this theorem is left to the reader as an exercise (see Exercise 18). Consider the asymptotic behavior of the ratio cn = |an+1 |/|an |
as n → ∞. The theorem asserts that if cn behaves as cn ∼ 1 + b/n for
2
large
P n (i.e. neglecting terms of order 1/n and higher), then the series
an converges if b < −1.
For a p−series the ratio test is inconclusive (see the proof of the
ratio test). However, De Mogran’s test resolves the inconclusiveness.
Indeed, for large n
cn =
1
p
np
=
∼ 1−
p
p
(n + 1)
(1 + 1/n)
n
where the asymptotic behavior has been found from the linearization
f(x) = (1 + x)−p ∼ f(0) + f 0 (0)x = 1 − px for small x = 1/n. So, here
b = −p and the series converges if b < −1 or p > 1.
This illustrates a basic technical trick to apply De Morgan’s test.
Suppose that there is a function f(x) such that |an+1 |/|an | = f(1/n).
If f is differentiable at x = 0, i.e., f(x) ≈ f(0) + f 0 (0)x as x → 0, then
|an+1 |/|an | = f(1/n) ∼ f(0) + f 0 (0)(1/n) = 1 + f 0 (0)/n
P
and the series
an converges absolutely if f 0 (0) < −1. Note that the
property f(0) = 1 follows from the inconclusiveness of the ratio test.
60.7. Exercises.
1-15. Determine whether the series is absolutely convergent, conditionally convergent, or divergent (here p is real)