18.
100A: Complete Lecture Notes
Lecture 12:
The Ratio, Root, and Alternating Series Tests
We continue our study of convergence tests.
Theorem 1 (Ratio test)
Suppose xn 6= 0 for all n and
|xn+1 |
L = lim
n→∞ |xn |
exists. Then,
P
1. if L < 1 then xn converges absolutely.
P
2. if L > 1 then xn diverges.
Proof : We will first prove the second part of this theorem.
|xn+1 |
2) Suppose L > 1 and α ∈ (1, L). Then, there exists M0 ∈ N such that for all N ≥ M0 , |xn | ≥ α ≥ 1. Thus,
for all n ≥ M0 ,
|xn+1 | ≥ |xn | =⇒ lim |xn | 6= 0.
n→∞
P
Therefore, xn diverges.
|xn+1 |
1) Now suppose that L < 1. Let α ∈ (L, 1). Then, there exists M0 ∈ N such that ∀n ≥ M0 , |xn | < α.
Therefore,∀n ≥ M0 , |xn+1 | ≤ α|xn |. In other words, for all n ≥ M0 ,
|xn | ≤ α|xn−1 | ≤ α2 |xn−2 | ≤ · · · ≤ αn−M0 |xM0 |.
Let m ∈ N. Then,
m
X M
X0 −1 m
X
|xn | = |xn | + |xn |
n=1 n=1 n=M0
M
X0 −1 m
X
≤ |xn | + |xM0 | αn−M0
n=1 n=M0
M
X0 −1 ∞
X
≤ |xn | + |xM0 | α`
n=1 `=0
M0 −1
X |xM0 |
= |xn | + .
n=1
1−α
Pm ∞ P
Therefore, { n=1 |xn |}m=1 is bounded, and thus |xn | converges. Hence, xn is absolutely convergent.
Let’s consider two examples where we can use the Ratio test.
1
Example 2
P∞ (−1)n
Show the series n=1 n2 +1 converges absolutely.
Proof : Notice
(−1)n 1 1
≤ 2 < 2,
n2 + 1 n +1 n
and hence
(−1)n+1
(n+1)2 +1 n2
lim (−1)n
< lim = 1.
n→∞ n→∞ (n + 1)2
n2 +1
Example 3
xn
P∞
Show that ∀x ∈ R, n=0 n! converges absolutely.
Proof : This immediately follows from the Ratio test, noting that
|x|n+1 n! |x|
lim · n = lim = 0.
n→∞ (n + 1)! |x| n→∞ n + 1
Remark 4. As seen above, the Ratio test can be really helpful to use when we have a (−1)n or a factorial in the
argument. Also note that if L = 1 then we the test doesn’t apply.
Theorem 5 (Root test)
P
Let xn be a series and suppose that
L = lim |xn |1/n
n→∞
exists. Then,
P
1. if L < 1 then xn converges absolutely.
P
2. if L > 1 then xn diverges.
Proof :
1. Suppose L < 1. Let L < r < 1. Then, since |xn |1/n → L, ∃M ∈ N such that ∀n ≥ M , |xn |1/n < r. Therefore,
for all n ≥ M , |xn | ≤ rn . Thus, for all m ∈ N,
m
X M
X −1 m
X
|xn | = |xn | + |xn |
n=1 n=1 n=M
M
X −1 m
X
≤ |xn | + rn
n=1 n=M
M
X −1 ∞
X
≤ |xn | + rn
n=1 n=M
M −1
X rM
= |xn | + .
n=1
1−r
Pm ∞ P
Thus, { n=1 |xn |}m=1 is bounded, and thus |xn | converges.
2
2. Suppose L > 1. Then, since |xn |1/n → L > 1, there exists an M ∈ N such that for all n ≥ M , |xn |1/n > 1.
P
In other words, for all n ≥ M , |xn | > 1. Therefore, limn→∞ xn 6= 0, and thus xn diverges.
Remark 6. Again, note that if L = 1 then the test doesn’t apply.
Theorem 7 (Alternating Series test)
(−1)n xn converges.
P
Let {xn } be a monotone decreasing sequence such that xn → 0. Then,
Pm n
Proof : Let sm = n=1 (−1) xn . Then,
2k
X
s2k = (−1)n xn
n=1
= (x2 − x1 ) + (x4 − x3 ) + · · · + (x2k − x2k−1 )
≥ (x2 − x1 ) + · · · + (x2k − x2k−1 ) + (x2k+2 − x2k+1 )
= s2(k+1)
as {xn } is a monotone decreasing sequence. Thus, {s2k }∞
k=1 is monotone decreasing. Furthermore,
s2k = −x1 + (x2 − x3 ) + (x4 − x5 ) + · · · + (x2k−2 − x2k−1 ) + x2k ≥ −x1 .
In other words, {s2k } is a bounded below monotone decreasing sequence. Thus, {s2k }∞
k=1 converges. Let s =
limk→∞ s2k . We now prove {sm }∞m=1 converges to s.
Let > 0. Since s2k → s, ∃M0 ∈ N such that for all k ≥ M0 ,
|s2k − s| < .
2
Since xn → 0, ∃M1 ∈ N such that ∀n ≥ M1 ,
. |xn | <
2
m
Choose M = max{2M+ 0 + 1, M1 }. Suppose m ≥ M . If m is even, then 2 ≥ M0 + 1/2 ≥ M0 . Therefore,
|sm − s| = |s2· m2 − s| < < .
2
m−1
If m is odd, let k = 2 so m = 2k + 1. Then, m ≥ M =⇒ k ≥ M0 and m ≥ M1 . Then,
|sm − s| = |sm−1 + xm − s|
≤ |s2k − s + xm |
≤ |s2k − s| + |xm | < + = .
2 2
(−1)n xn converges.
P
Thus, sm → s, and thus
Corollary 8
P (−1)n P (−1)n
We already showed that n does not absolutely converge. However, n converges.
Proof : This follows immediately from the Alternating Series test.
3
Theorem 9
P P P
Suppose xn converges absolutely and xn = x. Let σ : N → N be a bijective function. Then, xσ(n)
P
is absolutely convergent and xσ(n) = x. In other words, absolute convergence implies if we rearrange the
sequence the new series will still converge to the same value of the original series.
P Pm
Proof : We first show |xσ(n) | converges, which is equivalent to showing the partial sums n=1 |xσ(n) | is bounded.
P
Since xn converges, ∃B ≥ 0 such that for all ` ∈ N,
`
X
|xn | ≤ B.
n=1
Let m ∈ N. Then, σ({1, . . . , m}) is a finite subset of N. Thus, there exists an ` ∈ N such that
σ({1, . . . , m}) ⊂ {1, . . . , `}.
Thus,
m
X X `
X
|xσ(n) | = |xn | ≤ |xn | ≤ B.
n=1 n∈σ({1,...,m}) n=1
P P∞
Therefore, |xσ(n) | converges. Let x = n=1 xn , and let > 0. Then, ∃M0 ∈ N such that ∀m ≥ M0 ,
m
X
xn − x < .
n=1
2
P
Since |xn | converges, ∃M1 ∈ N such that for all ` > m ≥ M1 ,
`
X
|xn | < .
n=m+1
2
Let M2 = max{M0 , M1 }. Then, ∀` > m ≥ M2 ,
m `
X X
xn − x < and |xn | < .
n=1
2 n=m+1
2
Since σ −1 ({1, . . . , M2 }) is a finite set, ∃M3 ∈ N such that
{1, . . . , M2 } ⊂ σ({1, . . . , M3 }).
4
Choose M = M3 . Thus, if m0 ≥ M ,
0
m
X X
xσ(n0 ) − x = xn − x
n0 =1 n∈σ({1,...,m0 })
M
X X
= xn − x + xn
n=1 n∈σ({1,...,m0 })\{1,...,M }
M max σ({1,...,m0 })
X X
≤ xn − x + |xn |
n=1 n=M +1
M
X `
X
≤ xn − x + |xn |
n=1 n=M +1
< + = .
2 2
5
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