3Convergence of infinite sums
IA Analysis I
3.3 Convergence tests
We’ll now come up with a lot of convergence tests.
Lemma (Alternating sequence test). Let (
a
n
) be a decreasing sequence of
non-negative reals, and suppose that
a
n
→
0. Then
∞
X
n=1
(
−
1)
n+1
a
n
converges,
i.e. a
1
− a
2
+ a
3
− a
4
+ ··· converges.
Proof. Let S
N
=
N
X
n=1
(−1)
n+1
a
n
. Then
S
2n
= (a
1
− a
2
) + (a
3
− a
4
) + ··· + (a
2n−1
− a
2n
) ≥ 0,
and (S
2n
) is an increasing sequence.
Also,
S
2n+1
= a
1
− (a
2
− a
3
) − (a
4
− a
5
) − ··· − (a
2n
− a
2n+1
),
and (S
2n+1
) is a decreasing sequence. Also S
2n+1
− S
2n
= a
2n+1
≥ 0.
Hence we obtain the bounds 0
≤ S
2n
≤ S
2n+1
≤ a
1
. It follows from the
monotone sequences property that (S
2n
) and (S
2n+1
) converge.
Since S
2n+1
− S
2n
= a
2n+1
→ 0, they converge to the same limit.
Example.
1 −
1
3
√
2
+
1
3
√
3
−
1
3
√
4
+ ···converges.
Lemma (Ratio test). We have three versions:
(i) If ∃c < 1 such that
|a
n+1
|
|a
n
|
≤ c,
for all n, then
P
a
n
converges.
(ii) If ∃c < 1 and ∃N such that
(∀n ≥ N)
|a
n+1
|
|a
n
|
≤ c,
then
P
a
n
converges. Note that just because the ratio is always less than
1, it doesn’t necessarily converge. It has to be always less than a fixed
number c. Otherwise the test will say that
P
1/n converges.
(iii) If ∃ρ ∈ (−1, 1) such that
a
n+1
a
n
→ ρ,
then
P
a
n
converges. Note that we have the open interval (
−
1
,
1). If
|a
n+1
|
|a
n
|
→ 1, then the test is inconclusive!
Proof.
(i) |a
n
| ≤ c
n−1
|a
1
|
. Since
P
c
n
converges, so does
P
|a
n
|
by comparison test.
So
P
a
n
converges absolutely, so it converges.
(ii)
For all
k ≥
0, we have
|a
N+k
| ≤ c
k
|a
N
|
. So the series
P
|a
N+k
|
converges,
and therefore so does
P
|a
k
|.
(iii)
If
a
n+1
a
n
→ ρ
, then
|a
n+1
|
|a
n
|
→ |ρ|
. So (setting
ε
= (1
− |ρ|
)
/
2) there exists
N
such that ∀n ≥ N,
|a
n+1
|
|a
n
|
≤
1+|ρ|
2
< 1. So the result follows from (ii).
Example. If |b| < 1, then
P
nb
n
converges, since
a
n+1
a
n
=
(n + 1)b
n+1
nb
n
=
1 +
1
n
b → b < 1.
So it converges.
We can also evaluate this directly by considering
∞
X
i=1
∞
X
n=i
b
n
.
The following two tests were taught at the end of the course, but are included
here for the sake of completeness.
Theorem (Condensation test). Let (
a
n
) be a decreasing non-negative sequence.
Then
P
∞
n=1
a
n
< ∞ if and only if
∞
X
k=1
2
k
a
2
k
< ∞.
Proof.
This is basically the proof that
P
1
n
diverges and
P
1
n
α
converges for
α < 1 but written in a more general way.
We have
a
1
+ a
2
+ (a
3
+ a
4
) + (a
5
+ ··· + a
8
) + (a
9
+ ··· + a
16
) + ···
≥a
1
+ a
2
+ 2a
4
+ 4a
8
+ 8a
16
+ ···
So if
P
2
k
a
2
k
diverges,
P
a
n
diverges.
To prove the other way round, simply group as
a
1
+ (a
2
+ a
3
) + (a
4
+ ··· + a
7
) + ···
≤a
1
+ 2a
2
+ 4a
4
+ ··· .
Example. If a
n
=
1
n
, then 2
k
a
2
k
= 1. So
P
∞
k=1
2
k
a
2
k
= ∞. So
P
∞
n=1
1
n
= ∞.
After we formally define integrals, we will prove the integral test:
Theorem (Integral test). Let
f
: [1
, ∞
]
→ R
be a decreasing non-negative
function. Then
P
∞
n=1
f(n) converges iff
R
∞
1
f(x) dx < ∞.