3Convergence of infinite sums
IA Analysis I
3.1 Infinite sums
Definition (Convergence of infinite sums and partial sums). Let (
a
n
) be a real
sequence. For each N, define
S
N
=
N
X
n=1
a
n
.
If the sequence (S
N
) converges to some limit s, then we say that
∞
X
n=1
a
n
= s,
and we say that the series
∞
X
n=1
a
n
converges.
We call S
N
the N th partial sum.
There is an immediate necessary condition for a series to converge.
Lemma. If
∞
X
n=1
a
n
converges. Then a
n
→ 0.
Proof.
Let
∞
X
n=1
a
n
=
s
. Then
S
n
→ s
and
S
n−1
→ s
. Then
a
n
=
S
n
− S
n−1
→
0.
However, the converse is false!
Example (Harmonic series). If a
n
= 1/n, then a
n
→ 0 but
P
a
n
= ∞.
We can prove this as follows:
S
2
n
− S
2
n−1
=
1
2
n−1
+ 1
+ ··· +
1
2
n
≥
2
n−1
2
n
=
1
2
.
Therefore S
2
n
≥ S
1
+ n/2. So the partial sums are unbounded.
Example (Geometric series). Let |ρ| < 1. Then
∞
X
n=0
ρ
n
=
1
1 −ρ
.
We can prove this by considering the partial sums:
N
X
n=0
ρ
n
=
1 −ρ
N+1
1 −ρ
.
Since ρ
N+1
→ 0, this tends to 1/(1 − ρ).
Example.
∞
X
n=2
1
n(n −1)
converges. This is since
1
n(n −1)
=
1
n −1
−
1
n
.
So
N
X
n=2
1
n(n −1)
= 1 −
1
N
→ 1.
Lemma. Suppose that
a
n
≥
0 for every
n
and the partial sums
S
n
are bounded
above. Then
P
∞
n=1
a
n
converges.
Proof.
The sequence (
S
n
) is increasing and bounded above. So the result follows
form the monotone sequences property.
The simplest convergence test we have is the comparison test. Roughly
speaking, it says that if 0
≤ a
n
≤ b
n
for all
n
and
P
b
n
converges, then
P
a
n
converges. However, we will prove a much more general form here for convenience.
Lemma (Comparison test). Let (
a
n
) and (
b
n
) be non-negative sequences, and
suppose that
∃C, N
such that
∀n ≥ N
,
a
n
≤ Cb
n
. Then if
P
b
n
converges, then
so does
P
a
n
.
Proof.
Let
M > N
. Also for each
R
, let
S
R
=
P
R
n=1
a
n
and
T
R
=
P
R
n=1
b
n
. We
want S
R
to be bounded above.
S
M
− S
N
=
M
X
n=N+1
a
n
≤ C
M
X
n=N+1
b
n
≤ C
∞
X
n=N+1
b
n
.
So
∀M ≥ N
,
S
M
≤ S
n
+
C
P
∞
n=N+1
b
n
. Since the
S
M
are increasing and
bounded, it must converge.
Example.
(i)
P
1
n2
n
converges, since
P
1
2
n
converges.
(ii)
P
n
2
n
converges.
If
n ≥
4, then
n ≤
2
n/2
. That’s because 4 = 2
4/2
and for
n ≥
4,
(
n
+ 1)
/n <
√
2
, so when we increase
n
, we multiply the right side by a
greater number by the left. Hence by the comparison test, it is sufficient
to show that
P
2
n/2
/
2
n
=
P
2
−n/2
converges, which it does (geometric
series).
(iii)
P
1
√
n
diverges, since
1
√
n
≥
1
n
. So if it converged, then so would
P
1
n
, but
P
1
n
diverges.
(iv)
P
1
n
2
converges, since for
n ≥
2,
1
n
2
≤
1
n(n−1)
, and we have proven that
the latter converges.
(v) Consider
∞
X
n=1
n + 5
n
3
− 7n
2
/2
. We show this converges by noting
n
3
−
7n
2
2
= n
2
n −
7
2
.
So if n ≥ 8, then
n
3
−
7n
2
2
≥
n
3
2
.
Also, n + 5 ≤ 2n. So
n + 5
n
3
− 7n
2
/2
≤ 4/n
2
.
So it converges by the comparison test.
(vi) If α > 1, then
P
1/n
α
converges.
Let S
N
=
P
N
n=1
1/n
α
. Then
S
2
n
− S
2
n−1
=
1
(2
n−1
+ 1)
α
+ ··· +
1
(2
n
)
α
≤
2
n−1
(2
n−1
)
α
= (2
n−1
)
1−α
= (2
1−α
)
n−1
.
But 2
1−α
< 1. So
S
2
n
= (S
2
n
− S
2
n−1
) + (S
2
n−1
− S
2
n−2
) + ···(S
2
− S
1
) + S
1
and is bounded above by comparison with the geometric series 1 + 2
1−α
+
(2
1−α
)
2
+ ···