The Omnibus Theorem of (Main) Convergence Tests
Definition 1 Let F be either IR or C. Let an n∈IN be a sequence in F .
(a) The sequence an n∈IN converges to A ∈ F if
∀ε > 0 ∃N ∈ IN s.t. an − A < ε whenever n ≥ N
A sequence which does not converge to any A is said to diverge.
(b) The sequence an n∈IN is said to be Cauchy if
∀ε > 0 ∃N ∈ IN s.t. am − an < ε whenever m, n ≥ N
P∞
(c) The series n=1 an is said to converge to S if the sequence sn n∈IN of partial sums
n
P
sn =
ak converges to S. A series which does not converge to any S is said to diverge.
k=1
(d) The series
P∞
n=1
an is said to converge absolutely if the
P∞
n=1
|an | converges.
Theorem 2 Let an n∈IN , cn n∈IN and dn n∈IN be sequences of real or complex numbers
n
P
and sn =
ak . Then
k=1
(a) (Cauchy Criterion)
∞
X
k=1
m
X
ak < ε whenever m ≥ n ≥ N
ak converges ⇐⇒ ∀ε > 0 ∃N ∈ IN s.t. k=n
(b) (Divergence Test)
∞
P
an converges =⇒ lim an = 0
n→∞
n=1
(c) If an ≥ 0 for all n ∈ IN, then
∞
X
ak converges ⇐⇒
k=1
sn
n∈IN
is bounded
(d) absolute convergence =⇒ convergence
(e) (Comparison Test) Let N0 ∈ IN.
∞
∞
P
P
(i) |an | ≤ cn ∀n ≥ N0 and
cn converges =⇒
an converges absolutely
n=1
(ii) an ≥ dn ≥ 0 ∀n ≥ N0 and
∞
P
n=1
∞
P
dn diverges =⇒
c Joel Feldman.
2014. All rights reserved.
an diverges
n=1
n=1
October 8, 2014
(Main) Convergence Tests
1
(f ) (Integral Test) Let c ∈ IR. If f (x) is a real valued function which is defined and
continuous for all x ≥ c and which obeys
(i) f (x) ≥ 0 ∀x ≥ c
(ii) f (x) is monotonically decreasing, i.e. y > x =⇒ f (y) ≤ f (x)
(iii) f (n) = an ∀n ≥ c
then
y = f (x)
y
Z ∞
a1
∞
X
a2
an converges ⇐⇒
f (x) dx < ∞
a3
a4
c
n=1
Furthermore, if N ≥ c, the truncation error
Z
∞
X
≤
a
−
s
n
N
n=1
1
2
3
4 x
∞
f (x) dx
N
(g) (Examples)
∞
P
(i) (Geometric Series)
(ii) (p Test)
(iii)
∞
P
n=2
∞
P
n=1
1
n(ln n)p
r n converges to
n=0
1
np
1
1−r
if |r| < 1 and diverges if |r| ≥ 1.
converges if p > 1 and diverges if p ≤ 1.
converges if p > 1 and diverges if p ≤ 1.
(h) (Root Test) Let α = lim sup
n→∞
p
n
|an |. Then
α < 1 =⇒
α > 1 =⇒
∞
X
n=1
∞
X
an converges absolutely
an diverges
n=1
α = 1 =⇒ nothing
(i) (Ratio Test) If an
∞
X
n=1
n∈IN
is a sequence of nonzero numbers, then
a an converges absolutely if lim sup n+1
an < 1
c Joel Feldman.
n→∞
an+1 diverges if an ≥ 1 for all n bigger than some natural number N
2014. All rights reserved.
October 8, 2014
(Main) Convergence Tests
2
(j) (Alternating Series Test) If
(i) an ≥ 0 ∀n ∈ IN
(ii) an+1 ≤ an ∀n ∈ IN
(iii) limn→∞ an = 0
∞
P
then s =
(−1)n−1 an converges and s − sn is between 0 and (the first dropped term)
n=1
(−1)n an+1 .
Proof:
(a) ⇒
Let lim sn = s and let ε > 0. Then
n→∞
∃Ñ ∈ IN s.t. n ≥ Ñ =⇒ |sn − s| <
ε
2
Set N = Ñ + 1. Then
m ≥ n ≥ N =⇒ m > n − 1 ≥ Ñ =⇒ |sm − s| < 2ε , |sn−1 − s| <
ε
2
=⇒ |sm − sn−1 | = |(sm − s) + (s − sn−1 )| < ε
m
X ak < ε
=⇒ k=n
(a) ⇐
By hypothesis
∀ε > 0 ∃Nε ∈ IN s.t. m ≥ n ≥ Nε =⇒ |sm − sn | < ε
=⇒ sn n∈IN is bounded
(Choosing ε = 1, n = N1 , we see that sm m ≥ N1 is bounded.)
=⇒ sn n∈IN has a subsequential limit. Say lim sin = s.
n→∞
=⇒ ∀ε > 0 n ≥ Nε/2 =⇒ |sn − s| = sn − sim + sim − s < 2ε + 2ε = ε
if we choose m large enough that im ≥ Nε/2 and sim − s < 2ε .
(b) Apply part (a) with m = n.
(c) As an ≥ 0 for all n ∈ IN, the sequence sn
converges if and only if it is bounded.
(d) If
∞
P
k=1
the series
|ak | converges, then
∞
P
∞
P
k=1
n∈IN
is monotonically increasing. So it
P
m
P
m
|ak |,
ak ≤
|ak | obeys the Cauchy criterion. As k=n
k=n
ak then also obeys the Cauchy criterion and hence converges.
k=1
c Joel Feldman.
2014. All rights reserved.
October 8, 2014
(Main) Convergence Tests
3
(e.i) The sequence
n
o
N0
n
∞
P
P
P
tn =
|ak | is monotonic and bounded above by
|ak | +
ck
k=1
and hence converges. So
∞
P
k=1
k=1
ak converges absolutely and hence also converges by part (d).
k=1
∞
P
(e.ii) If
∞
P
ak were to converge, then
k=1
dk would also converge by part (e.i). This is a
k=1
contradiction.
∞
P
(f) Let N be any fixed integer bigger than c + 1. Then
∞
P
ak converges if and only if
k=1
ak converges, which in turn is the case if and only if the sequence tn =
k=N
n
P
ak is
k=N
bounded, since ak ≥ 0 for all k ≥ N > c. As f (x) decreases as x increases, we have that
f (x) ≥ ak when x ≤ k and f (x) ≤ ak when x ≥ k. Thus, for any k ≥ N
Z
k+1
f (x) dx ≤ ak ≤
Z
k
f (x) dx
k−1
k
Summing k from N to n,
Z
n+1
f (x) dx ≤
N
ak
n
X
ak ≤
Z
y = f (x)
n
f (x) dx
N−1
k=N
So the tn ’s are bounded if and only if
(g,i) Exercise.
(g.ii) If p ≤ 0, the series
∞
P
n=1
1
np
=
R∞
c
∞
P
k−1
k+1
k
f (x) dx < ∞.
n|p| diverges by part (b). The remaining cases
n=1
are easily proven using the integral test with c = 1 and f (x) = x1p , which is positive and
monotonically decreasing for p > 0. We just have to observe that
Z L
ln L
if p = 1
1
L1−p −1
xp dx =
if p 6= 1
1−p
1
1
when p > 1.
grows unboundedly as L → ∞ when p ≤ 1 and converges to p−1
Here is a second proof that does not use integrals. If p > 0, each successive term in
the series is smaller than the term before. So each term with index n between 2m and
2m+1 obeys
1
1
≤ n1p ≤ 2mp
2(m+1)p
There are 2m+1 − 2m such terms. Hence
1 (m+1)(1−p)
2
2
=
2m+1 −2m
2(m+1)p
≤
X
1
np
≤
2m+1 −2m
2mp
=
1
2m(p−1)
2m ≤n<2m+1
c Joel Feldman.
2014. All rights reserved.
October 8, 2014
(Main) Convergence Tests
4
If 0 ≤ p ≤ 1, the left hand bound 21 2(m+1)(1−p) ≥ 12 . Summing m from 0 to N − 1, we
2N
−1
P
1
N
have
np ≥ 2 which forces divergence. If p > 1, then summing m from 0 to N − 1
n=1
in the right hand bound, we have
N
2
P
n=1
1
np
≤
N−1
P
m=0
bounded and hence converges by part (c).
m
1
2p−1
≤ 1−
−1
1
.
2p−1
So the series is
|p|
(g.iii) If p ≤ 0, then for all n ≥ 2, n(ln1n)p ≥ (ln n2) and the series diverges by the
comparison test. Again, the remaining cases are easily proven using the integral test, this
time with c = 2 and f (x) = x(ln1x)p . To compute the integral, we make the change of
variables y = ln x, dy = dx
x .
Z
L
2
1
x(ln x)p
dx =
Z
ln L
ln 2
1
yp
(
)
ln(ln L) − ln(ln 2) if p = 1
dy = (ln L)1−p −(ln 2)1−p
if p 6= 1
1−p
∞
if p ≤ 1
1−p
as L → ∞
→ (ln 2)
if p > 1
p−1
Again, here is a second proof that does not use the integral test. If p > 0, each term
with index n between 2m and 2m+1 obeys
1
2m+1 (m+1)p (ln 2)p
≤
1
n(ln n)p
≤
1
2m (m ln 2)p
Hence
1
1
2(ln 2)p (m+1)p
=
2m+1 −2m
2m+1 (m+1)p (ln 2)p
≤
X
1
n(ln n)p
≤
2m+1 −2m
2m (m ln 2)p
=
1
1
(ln 2)p mp
2m ≤n<2m+1
Summing m from 1 to N − 1, we have
1
2(ln 2)p
N−1
X
1
(m+1)p
m=1
≤
N
2X
−1
1
n(ln n)p
≤
n=2
1
(ln 2)p
N−1
X
1
mp
m=1
So if p > 1, the right hand inequality together with part (g.ii) shows that the series is
bounded and hence converges, by part (c), and if p ≤ 1, the left hand inequality together
with part (g.ii) shows that the series diverges.
(h) If α < 1, pick any r obeying α < r < 1. There is a natural number N such that
p
n
|an | < r, or equivalently, |an | < r n , for all n ≥ N . So convergence follows by comparison
with the geometric series. On the other hand, if α > 1, there must be a subsequence with
p
nk
|ank | ≥ 1, or equivalently, |ank | ≥ 1. So the series diverges by part (b).
c Joel Feldman.
2014. All rights reserved.
October 8, 2014
(Main) Convergence Tests
5
a ≥ 1 for all n ≥ N , then |an | ≥ |aN | > 0 for all n ≥ N and the series diverges
(i) If an+1
n
< 1, there is a r < 1 and an N ∈ IN such
by part (b). On the other hand, if lim sup an+1
an
n→∞
≤ r for all n ≥ N . Consequently |aN+1 | ≤ r|aN |, |aN+2 | ≤ r|aN+1 | ≤ r 2 |aN |
that aan+1
n
and so on. Iterating gives |aN+p | ≤ r p |aN | which implies convergence by comparison with
the geometric series.
(j) Exercise
c Joel Feldman.
2014. All rights reserved.
October 8, 2014
(Main) Convergence Tests
6