In this chapter we will see an easy and useful criterion for divergence. This criterion is called term test or also null-sequence test or divergence test . It says that every series
∑
k
=
1
∞
a
k
{\displaystyle \sum _{k=1}^{\infty }a_{k}}
, where
(
a
k
)
k
∈
N
{\displaystyle (a_{k})_{k\in \mathbb {N} }}
is not a null sequence, is divergent. When you invert that, this means that for every convergent series
∑
k
=
1
∞
a
k
{\displaystyle \sum _{k=1}^{\infty }a_{k}}
we have that
lim
k
→
∞
a
k
=
0
{\displaystyle \lim _{k\to \infty }a_{k}=0}
(
(
a
k
)
k
∈
N
{\displaystyle (a_{k})_{k\in \mathbb {N} }}
is a null sequence).
Proof (Term test)
Our premise is that
∑
k
=
1
∞
a
k
{\displaystyle \sum _{k=1}^{\infty }a_{k}}
is a convergent series. We want to conclude that
(
a
n
)
n
∈
N
{\displaystyle (a_{n})_{n\in \mathbb {N} }}
is a null sequence.
A member of the sequence
(
a
n
)
n
∈
N
{\displaystyle (a_{n})_{n\in \mathbb {N} }}
can be written as the difference between two consecutive partial sums
S
n
=
∑
k
=
1
n
a
k
{\displaystyle S_{n}=\sum _{k=1}^{n}a_{k}}
and
S
n
−
1
=
∑
k
=
1
n
−
1
a
k
{\displaystyle S_{n-1}=\sum _{k=1}^{n-1}a_{k}}
:
a
n
=
a
n
+
0
=
a
n
+
a
n
−
1
+
a
n
−
2
+
…
+
a
2
+
a
1
−
a
n
−
1
−
a
n
−
2
−
…
−
a
2
−
a
1
=
(
a
n
+
a
n
−
1
+
a
n
−
2
+
…
+
a
2
+
a
1
)
−
(
a
n
−
1
+
a
n
−
2
+
…
+
a
2
+
a
1
)
=
∑
k
=
1
n
a
k
−
∑
k
=
1
n
−
1
a
k
=
S
n
−
S
n
−
1
{\displaystyle {\begin{aligned}a_{n}&=a_{n}+0\\&=a_{n}+a_{n-1}+a_{n-2}+\ldots +a_{2}+a_{1}-a_{n-1}-a_{n-2}-\ldots -a_{2}-a_{1}\\&={\color {Teal}(a_{n}+a_{n-1}+a_{n-2}+\ldots +a_{2}+a_{1})}-{\color {Indigo}(a_{n-1}+a_{n-2}+\ldots +a_{2}+a_{1})}\\&={\color {Teal}\sum _{k=1}^{n}a_{k}}-{\color {Indigo}\sum _{k=1}^{n-1}a_{k}}=S_{n}-S_{n-1}\end{aligned}}}
From our premise we know that
∑
k
=
1
∞
a
k
{\displaystyle \sum _{k=1}^{\infty }a_{k}}
has a limit
s
∈
R
{\displaystyle s\in \mathbb {R} }
. Thus we have
lim
n
→
∞
S
n
=
lim
n
→
∞
∑
k
=
1
n
a
k
=
s
{\displaystyle \lim _{n\to \infty }S_{n}=\lim _{n\to \infty }\sum _{k=1}^{n}a_{k}=s}
But also
lim
n
→
∞
S
n
−
1
=
s
{\displaystyle \lim _{n\to \infty }S_{n-1}=s}
, because the limit will not change if we simply shift indexes. Put together we obtain:
lim
n
→
∞
a
n
↓
a
n
=
S
n
−
S
n
−
1
(see above)
=
lim
n
→
∞
(
S
n
−
S
n
−
1
)
↓
lim
n
→
∞
(
a
n
−
b
n
)
=
lim
n
→
∞
a
n
−
lim
n
→
∞
b
n
=
lim
n
→
∞
S
n
−
lim
n
→
∞
S
n
−
1
=
s
−
s
=
0
{\displaystyle {\begin{aligned}&\lim _{n\to \infty }a_{n}\\[0.3em]&\quad {\color {OliveGreen}\left\downarrow \ a_{n}=S_{n}-S_{n-1}{\text{ (see above)}}\right.}\\[0.3em]=&\lim _{n\to \infty }(S_{n}-S_{n-1})\\[0.3em]&\quad {\color {OliveGreen}\left\downarrow \ \lim _{n\to \infty }(a_{n}-b_{n})=\lim _{n\to \infty }a_{n}-\lim _{n\to \infty }b_{n}\right.}\\[0.3em]=&\lim _{n\to \infty }S_{n}-\lim _{n\to \infty }S_{n-1}=s-s=0\end{aligned}}}
We can conclude that
(
a
n
)
n
∈
N
{\displaystyle (a_{n})_{n\in \mathbb {N} }}
must be null sequence.
Proof (Term test)
We can proof the same result using the Cauchy criterion. As a reminder, every convergent series
∑
k
=
1
∞
a
k
{\displaystyle \sum _{k=1}^{\infty }a_{k}}
satisfies the Cauchy criterion:
∀
ϵ
>
0
∃
N
∈
N
∀
n
≥
m
≥
N
:
|
∑
k
=
m
n
a
k
|
<
ϵ
{\displaystyle \forall \epsilon >0\,\exists N\in \mathbb {N} \,\forall n\geq m\geq N:\left|\sum _{k=m}^{n}a_{k}\right|<\epsilon }
We don't consider all
n
≥
m
{\displaystyle n\geq m}
, but only the case
m
=
n
{\displaystyle m=n}
:
∀
ϵ
>
0
∃
N
∈
N
∀
n
=
m
≥
N
:
|
∑
k
=
m
n
a
k
|
<
ϵ
↓
n
=
m
⇒
∀
ϵ
>
0
∃
N
∈
N
∀
n
≥
N
:
|
∑
k
=
n
n
a
k
|
<
ϵ
↓
∑
k
=
n
n
a
k
=
a
n
⇒
∀
ϵ
>
0
∃
N
∈
N
∀
n
≥
N
:
|
a
n
|
<
ϵ
↓
|
a
n
|
=
|
a
n
−
0
|
⇒
∀
ϵ
>
0
∃
N
∈
N
∀
n
≥
N
:
|
a
n
−
0
|
<
ϵ
{\displaystyle {\begin{aligned}&\forall \epsilon >0\,\exists N\in \mathbb {N} \,\forall n=m\geq N:\left|\sum _{k=m}^{n}a_{k}\right|<\epsilon \\&\quad {\color {OliveGreen}\left\downarrow \ n=m\right.}\\\Rightarrow \ &\forall \epsilon >0\,\exists N\in \mathbb {N} \,\forall n\geq N:\left|\sum _{k=n}^{n}a_{k}\right|<\epsilon \\&\quad {\color {OliveGreen}\left\downarrow \ \sum _{k=n}^{n}a_{k}=a_{n}\right.}\\[0.5em]\Rightarrow \ &\forall \epsilon >0\,\exists N\in \mathbb {N} \,\forall n\geq N:\left|a_{n}\right|<\epsilon \\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ |a_{n}|=|a_{n}-0|\right.}\\[0.5em]\Rightarrow \ &\forall \epsilon >0\,\exists N\in \mathbb {N} \,\forall n\geq N:\left|a_{n}-0\right|<\epsilon \end{aligned}}}
The last formula is exactly the
ϵ
{\displaystyle \epsilon }
-definition for what it means that
(
a
n
)
n
∈
N
{\displaystyle (a_{n})_{n\in \mathbb {N} }}
is a null sequences. In other words we have showed that
lim
n
→
∞
a
n
=
0
{\displaystyle \lim _{n\to \infty }a_{n}=0}
.
Example exercise for the term test (in German)
Math for Non-Geeks: Template:Aufgabe
If we demand that
(
a
n
)
n
∈
N
{\displaystyle (a_{n})_{n\in \mathbb {N} }}
is monotonically decreasing, then we can show that even
(
n
a
n
)
n
∈
N
{\displaystyle (na_{n})_{n\in \mathbb {N} }}
is a null sequence. See the respective exercise .