Test for infinite series of monotonous terms for convergence
The integral test applied to the
harmonic series . Since the area under the curve y = 1/x for x ∈ [1, ∞) is infinite, the total area of the rectangles must be infinite as well.
In
mathematics , the integral test for convergence is a
method used to test infinite
series of
monotonous terms for
convergence . It was developed by
Colin Maclaurin and
Augustin-Louis Cauchy and is sometimes known as the Maclaurin–Cauchy test .
Statement of the test
Consider an
integer N and a function f defined on the unbounded
interval [N , ∞) , on which it is
monotone decreasing . Then the infinite series
∑
n
=
N
∞
f
(
n
)
{\displaystyle \sum _{n=N}^{\infty }f(n)}
converges to a
real number if and only if the
improper integral
∫
N
∞
f
(
x
)
d
x
{\displaystyle \int _{N}^{\infty }f(x)\,dx}
is finite. In particular, if the integral diverges, then the
series diverges as well.
If the improper integral is finite, then the proof also gives the
lower and upper bounds
∫
N
∞
f
(
x
)
d
x
≤
∑
n
=
N
∞
f
(
n
)
≤
f
(
N
)
+
∫
N
∞
f
(
x
)
d
x
{\displaystyle \int _{N}^{\infty }f(x)\,dx\leq \sum _{n=N}^{\infty }f(n)\leq f(N)+\int _{N}^{\infty }f(x)\,dx}
(1 )
for the infinite series.
Note that if the function
f
(
x
)
{\displaystyle f(x)}
is increasing, then the function
−
f
(
x
)
{\displaystyle -f(x)}
is decreasing and the above theorem applies.
Proof
The proof basically uses the
comparison test , comparing the term f (n ) with the integral of f over the intervals
[n − 1, n ) and [n , n + 1) , respectively.
The monotonous function
f
{\displaystyle f}
is
continuous
almost everywhere . To show this, let
D
=
{
x
∈
N
,
∞
)
∣
f
is discontinuous at
x
}
{\displaystyle D=\{x\in [N,\infty )\mid f{\text{ is discontinuous at }}x\}}
. For every
x
∈
D
{\displaystyle x\in D}
, there exists by the
density of
Q
{\displaystyle \mathbb {Q} }
a
c
(
x
)
∈
Q
{\displaystyle c(x)\in \mathbb {Q} }
so that
c
(
x
)
∈
lim
y
↓
x
f
(
y
)
,
lim
y
↑
x
f
(
y
)
{\displaystyle c(x)\in \left[\lim _{y\downarrow x}f(y),\lim _{y\uparrow x}f(y)\right]}
. Note that this set contains an
open
non-empty interval precisely if
f
{\displaystyle f}
is
discontinuous at
x
{\displaystyle x}
. We can uniquely identify
c
(
x
)
{\displaystyle c(x)}
as the
rational number that has the least index in an
enumeration
N
→
Q
{\displaystyle \mathbb {N} \to \mathbb {Q} }
and satisfies the above property. Since
f
{\displaystyle f}
is
monotone , this defines an
injective
mapping
c
:
D
→
Q
,
x
↦
c
(
x
)
{\displaystyle c:D\to \mathbb {Q} ,x\mapsto c(x)}
and thus
D
{\displaystyle D}
is
countable . It follows that
f
{\displaystyle f}
is
continuous
almost everywhere . This is
sufficient for
Riemann integrability .
[1]
Since f is a monotone decreasing function, we know that
f
(
x
)
≤
f
(
n
)
for all
x
∈
n
,
∞
)
{\displaystyle f(x)\leq f(n)\quad {\text{for all }}x\in [n,\infty )}
and
f
(
n
)
≤
f
(
x
)
for all
x
∈
N
,
n
.
{\displaystyle f(n)\leq f(x)\quad {\text{for all }}x\in [N,n].}
Hence, for every integer n ≥ N ,
∫
n
n
+
1
f
(
x
)
d
x
≤
∫
n
n
+
1
f
(
n
)
d
x
=
f
(
n
)
{\displaystyle \int _{n}^{n+1}f(x)\,dx\leq \int _{n}^{n+1}f(n)\,dx=f(n)}
(2 )
and, for every integer n ≥ N + 1 ,
f
(
n
)
=
∫
n
−
1
n
f
(
n
)
d
x
≤
∫
n
−
1
n
f
(
x
)
d
x
.
{\displaystyle f(n)=\int _{n-1}^{n}f(n)\,dx\leq \int _{n-1}^{n}f(x)\,dx.}
(3 )
By summation over all n from N to some larger integer M , we get from (
2 )
∫
N
M
+
1
f
(
x
)
d
x
=
∑
n
=
N
M
∫
n
n
+
1
f
(
x
)
d
x
⏟
≤
f
(
n
)
≤
∑
n
=
N
M
f
(
n
)
{\displaystyle \int _{N}^{M+1}f(x)\,dx=\sum _{n=N}^{M}\underbrace {\int _{n}^{n+1}f(x)\,dx} _{\leq \,f(n)}\leq \sum _{n=N}^{M}f(n)}
and from (
3 )
∑
n
=
N
M
f
(
n
)
=
f
(
N
)
+
∑
n
=
N
+
1
M
f
(
n
)
≤
f
(
N
)
+
∑
n
=
N
+
1
M
∫
n
−
1
n
f
(
x
)
d
x
⏟
≥
f
(
n
)
=
f
(
N
)
+
∫
N
M
f
(
x
)
d
x
.
{\displaystyle \sum _{n=N}^{M}f(n)=f(N)+\sum _{n=N+1}^{M}f(n)\leq f(N)+\sum _{n=N+1}^{M}\underbrace {\int _{n-1}^{n}f(x)\,dx} _{\geq \,f(n)}=f(N)+\int _{N}^{M}f(x)\,dx.}
Combining these two estimates yields
∫
N
M
+
1
f
(
x
)
d
x
≤
∑
n
=
N
M
f
(
n
)
≤
f
(
N
)
+
∫
N
M
f
(
x
)
d
x
.
{\displaystyle \int _{N}^{M+1}f(x)\,dx\leq \sum _{n=N}^{M}f(n)\leq f(N)+\int _{N}^{M}f(x)\,dx.}
Letting M tend to infinity, the bounds in (
1 ) and the result follow.
Applications
The
harmonic series
∑
n
=
1
∞
1
n
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n}}}
diverges because, using the
natural logarithm , its
antiderivative , and the
fundamental theorem of calculus , we get
∫
1
M
1
n
d
n
=
ln
n
|
1
M
=
ln
M
→
∞
for
M
→
∞
.
{\displaystyle \int _{1}^{M}{\frac {1}{n}}\,dn=\ln n{\Bigr |}_{1}^{M}=\ln M\to \infty \quad {\text{for }}M\to \infty .}
On the other hand, the series
ζ
(
1
+
ε
)
=
∑
n
=
1
∞
1
n
1
+
ε
{\displaystyle \zeta (1+\varepsilon )=\sum _{n=1}^{\infty }{\frac {1}{n^{1+\varepsilon }}}}
(cf.
Riemann zeta function )
converges for every ε > 0 , because by the
power rule
∫
1
M
1
n
1
+
ε
d
n
=
−
1
ε
n
ε
|
1
M
=
1
ε
(
1
−
1
M
ε
)
≤
1
ε
<
∞
for all
M
≥
1.
{\displaystyle \int _{1}^{M}{\frac {1}{n^{1+\varepsilon }}}\,dn=\left.-{\frac {1}{\varepsilon n^{\varepsilon }}}\right|_{1}^{M}={\frac {1}{\varepsilon }}\left(1-{\frac {1}{M^{\varepsilon }}}\right)\leq {\frac {1}{\varepsilon }}<\infty \quad {\text{for all }}M\geq 1.}
From (
1 ) we get the upper estimate
ζ
(
1
+
ε
)
=
∑
x
=
1
∞
1
n
1
+
ε
≤
1
+
ε
ε
,
{\displaystyle \zeta (1+\varepsilon )=\sum _{x=1}^{\infty }{\frac {1}{n^{1+\varepsilon }}}\leq {\frac {1+\varepsilon }{\varepsilon }},}
which can be compared with some of the
particular values of Riemann zeta function .
Borderline between divergence and convergence
The above examples involving the harmonic series raise the question of whether there are monotone sequences such that f (n ) decreases to 0 faster than 1/n but slower than 1/n 1+ε in the sense that
lim
n
→
∞
f
(
n
)
1
/
n
=
0
and
lim
n
→
∞
f
(
n
)
1
/
n
1
+
ε
=
∞
{\displaystyle \lim _{n\to \infty }{\frac {f(n)}{1/n}}=0\quad {\text{and}}\quad \lim _{n\to \infty }{\frac {f(n)}{1/n^{1+\varepsilon }}}=\infty }
for every ε > 0 , and whether the corresponding series of the f (n ) still diverges. Once such a sequence is found, a similar question can be asked with f (n ) taking the role of 1/n , and so on. In this way it is possible to investigate the borderline between divergence and convergence of infinite series.
Using the integral test for convergence, one can show (see below) that, for every
natural number k , the series
∑
n
=
N
k
∞
1
n
ln
(
n
)
ln
2
(
n
)
⋯
ln
k
−
1
(
n
)
ln
k
(
n
)
{\displaystyle \sum _{n=N_{k}}^{\infty }{\frac {1}{n\ln(n)\ln _{2}(n)\cdots \ln _{k-1}(n)\ln _{k}(n)}}}
(4 )
still diverges (cf.
proof that the sum of the reciprocals of the primes diverges for k = 1 ) but
∑
n
=
N
k
∞
1
n
ln
(
n
)
ln
2
(
n
)
⋯
ln
k
−
1
(
n
)
(
ln
k
(
n
)
)
1
+
ε
{\displaystyle \sum _{n=N_{k}}^{\infty }{\frac {1}{n\ln(n)\ln _{2}(n)\cdots \ln _{k-1}(n)(\ln _{k}(n))^{1+\varepsilon }}}}
(5 )
converges for every ε > 0 . Here lnk denotes the k -fold
composition of the natural logarithm defined
recursively by
ln
k
(
x
)
=
{
ln
(
x
)
for
k
=
1
,
ln
(
ln
k
−
1
(
x
)
)
for
k
≥
2.
{\displaystyle \ln _{k}(x)={\begin{cases}\ln(x)&{\text{for }}k=1,\\\ln(\ln _{k-1}(x))&{\text{for }}k\geq 2.\end{cases}}}
Furthermore, N k denotes the smallest natural number such that the k -fold composition is well-defined and lnk (N k ) ≥ 1 , i.e.
N
k
≥
e
e
⋅
⋅
e
⏟
k
e
′
s
=
e
↑↑
k
{\displaystyle N_{k}\geq \underbrace {e^{e^{\cdot ^{\cdot ^{e}}}}} _{k\ e'{\text{s}}}=e\uparrow \uparrow k}
using
tetration or
Knuth's up-arrow notation .
To see the divergence of the series (
4 ) using the integral test, note that by repeated application of the
chain rule
d
d
x
ln
k
+
1
(
x
)
=
d
d
x
ln
(
ln
k
(
x
)
)
=
1
ln
k
(
x
)
d
d
x
ln
k
(
x
)
=
⋯
=
1
x
ln
(
x
)
⋯
ln
k
(
x
)
,
{\displaystyle {\frac {d}{dx}}\ln _{k+1}(x)={\frac {d}{dx}}\ln(\ln _{k}(x))={\frac {1}{\ln _{k}(x)}}{\frac {d}{dx}}\ln _{k}(x)=\cdots ={\frac {1}{x\ln(x)\cdots \ln _{k}(x)}},}
hence
∫
N
k
∞
d
x
x
ln
(
x
)
⋯
ln
k
(
x
)
=
ln
k
+
1
(
x
)
|
N
k
∞
=
∞
.
{\displaystyle \int _{N_{k}}^{\infty }{\frac {dx}{x\ln(x)\cdots \ln _{k}(x)}}=\ln _{k+1}(x){\bigr |}_{N_{k}}^{\infty }=\infty .}
To see the convergence of the series (
5 ), note that by the
power rule , the chain rule and the above result
−
d
d
x
1
ε
(
ln
k
(
x
)
)
ε
=
1
(
ln
k
(
x
)
)
1
+
ε
d
d
x
ln
k
(
x
)
=
⋯
=
1
x
ln
(
x
)
⋯
ln
k
−
1
(
x
)
(
ln
k
(
x
)
)
1
+
ε
,
{\displaystyle -{\frac {d}{dx}}{\frac {1}{\varepsilon (\ln _{k}(x))^{\varepsilon }}}={\frac {1}{(\ln _{k}(x))^{1+\varepsilon }}}{\frac {d}{dx}}\ln _{k}(x)=\cdots ={\frac {1}{x\ln(x)\cdots \ln _{k-1}(x)(\ln _{k}(x))^{1+\varepsilon }}},}
hence
∫
N
k
∞
d
x
x
ln
(
x
)
⋯
ln
k
−
1
(
x
)
(
ln
k
(
x
)
)
1
+
ε
=
−
1
ε
(
ln
k
(
x
)
)
ε
|
N
k
∞
<
∞
{\displaystyle \int _{N_{k}}^{\infty }{\frac {dx}{x\ln(x)\cdots \ln _{k-1}(x)(\ln _{k}(x))^{1+\varepsilon }}}=-{\frac {1}{\varepsilon (\ln _{k}(x))^{\varepsilon }}}{\biggr |}_{N_{k}}^{\infty }<\infty }
and (
1 ) gives bounds for the infinite series in (
5 ).
See also
References
Knopp, Konrad , "Infinite Sequences and Series",
Dover Publications , Inc., New York, 1956. (§ 3.3)
ISBN
0-486-60153-6
Whittaker, E. T., and Watson, G. N., A Course in Modern Analysis , fourth edition, Cambridge University Press, 1963. (§ 4.43)
ISBN
0-521-58807-3
Ferreira, Jaime Campos, Ed Calouste Gulbenkian, 1987,
ISBN
972-31-0179-3