- Source: Limit comparison test
In mathematics, the limit comparison test (LCT) (in contrast with the related direct comparison test) is a method of testing for the convergence of an infinite series.
Statement
Suppose that we have two series
Σ
n
a
n
{\displaystyle \Sigma _{n}a_{n}}
and
Σ
n
b
n
{\displaystyle \Sigma _{n}b_{n}}
with
a
n
≥
0
,
b
n
>
0
{\displaystyle a_{n}\geq 0,b_{n}>0}
for all
n
{\displaystyle n}
.
Then if
lim
n
→
∞
a
n
b
n
=
c
{\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c}
with
0
<
c
<
∞
{\displaystyle 0
, then either both series converge or both series diverge.
Proof
Because
lim
n
→
∞
a
n
b
n
=
c
{\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c}
we know that for every
ε
>
0
{\displaystyle \varepsilon >0}
there is a positive integer
n
0
{\displaystyle n_{0}}
such that for all
n
≥
n
0
{\displaystyle n\geq n_{0}}
we have that
|
a
n
b
n
−
c
|
<
ε
{\displaystyle \left|{\frac {a_{n}}{b_{n}}}-c\right|<\varepsilon }
, or equivalently
−
ε
<
a
n
b
n
−
c
<
ε
{\displaystyle -\varepsilon <{\frac {a_{n}}{b_{n}}}-c<\varepsilon }
c
−
ε
<
a
n
b
n
<
c
+
ε
{\displaystyle c-\varepsilon <{\frac {a_{n}}{b_{n}}}
(
c
−
ε
)
b
n
<
a
n
<
(
c
+
ε
)
b
n
{\displaystyle (c-\varepsilon )b_{n}
As
c
>
0
{\displaystyle c>0}
we can choose
ε
{\displaystyle \varepsilon }
to be sufficiently small such that
c
−
ε
{\displaystyle c-\varepsilon }
is positive.
So
b
n
<
1
c
−
ε
a
n
{\displaystyle b_{n}<{\frac {1}{c-\varepsilon }}a_{n}}
and by the direct comparison test, if
∑
n
a
n
{\displaystyle \sum _{n}a_{n}}
converges then so does
∑
n
b
n
{\displaystyle \sum _{n}b_{n}}
.
Similarly
a
n
<
(
c
+
ε
)
b
n
{\displaystyle a_{n}<(c+\varepsilon )b_{n}}
, so if
∑
n
a
n
{\displaystyle \sum _{n}a_{n}}
diverges, again by the direct comparison test, so does
∑
n
b
n
{\displaystyle \sum _{n}b_{n}}
.
That is, both series converge or both series diverge.
Example
We want to determine if the series
∑
n
=
1
∞
1
n
2
+
2
n
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}+2n}}}
converges. For this we compare it with the convergent series
∑
n
=
1
∞
1
n
2
=
π
2
6
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}}
As
lim
n
→
∞
1
n
2
+
2
n
n
2
1
=
1
>
0
{\displaystyle \lim _{n\to \infty }{\frac {1}{n^{2}+2n}}{\frac {n^{2}}{1}}=1>0}
we have that the original series also converges.
One-sided version
One can state a one-sided comparison test by using limit superior. Let
a
n
,
b
n
≥
0
{\displaystyle a_{n},b_{n}\geq 0}
for all
n
{\displaystyle n}
. Then if
lim sup
n
→
∞
a
n
b
n
=
c
{\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c}
with
0
≤
c
<
∞
{\displaystyle 0\leq c<\infty }
and
Σ
n
b
n
{\displaystyle \Sigma _{n}b_{n}}
converges, necessarily
Σ
n
a
n
{\displaystyle \Sigma _{n}a_{n}}
converges.
Example
Let
a
n
=
1
−
(
−
1
)
n
n
2
{\displaystyle a_{n}={\frac {1-(-1)^{n}}{n^{2}}}}
and
b
n
=
1
n
2
{\displaystyle b_{n}={\frac {1}{n^{2}}}}
for all natural numbers
n
{\displaystyle n}
. Now
lim
n
→
∞
a
n
b
n
=
lim
n
→
∞
(
1
−
(
−
1
)
n
)
{\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\lim _{n\to \infty }(1-(-1)^{n})}
does not exist, so we cannot apply the standard comparison test. However,
lim sup
n
→
∞
a
n
b
n
=
lim sup
n
→
∞
(
1
−
(
−
1
)
n
)
=
2
∈
[
0
,
∞
)
{\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\limsup _{n\to \infty }(1-(-1)^{n})=2\in [0,\infty )}
and since
∑
n
=
1
∞
1
n
2
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}}
converges, the one-sided comparison test implies that
∑
n
=
1
∞
1
−
(
−
1
)
n
n
2
{\displaystyle \sum _{n=1}^{\infty }{\frac {1-(-1)^{n}}{n^{2}}}}
converges.
Converse of the one-sided comparison test
Let
a
n
,
b
n
≥
0
{\displaystyle a_{n},b_{n}\geq 0}
for all
n
{\displaystyle n}
. If
Σ
n
a
n
{\displaystyle \Sigma _{n}a_{n}}
diverges and
Σ
n
b
n
{\displaystyle \Sigma _{n}b_{n}}
converges, then necessarily
lim sup
n
→
∞
a
n
b
n
=
∞
{\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\infty }
, that is,
lim inf
n
→
∞
b
n
a
n
=
0
{\displaystyle \liminf _{n\to \infty }{\frac {b_{n}}{a_{n}}}=0}
. The essential content here is that in some sense the numbers
a
n
{\displaystyle a_{n}}
are larger than the numbers
b
n
{\displaystyle b_{n}}
.
Example
Let
f
(
z
)
=
∑
n
=
0
∞
a
n
z
n
{\displaystyle f(z)=\sum _{n=0}^{\infty }a_{n}z^{n}}
be analytic in the unit disc
D
=
{
z
∈
C
:
|
z
|
<
1
}
{\displaystyle D=\{z\in \mathbb {C} :|z|<1\}}
and have image of finite area. By Parseval's formula the area of the image of
f
{\displaystyle f}
is proportional to
∑
n
=
1
∞
n
|
a
n
|
2
{\displaystyle \sum _{n=1}^{\infty }n|a_{n}|^{2}}
. Moreover,
∑
n
=
1
∞
1
/
n
{\displaystyle \sum _{n=1}^{\infty }1/n}
diverges. Therefore, by the converse of the comparison test, we have
lim inf
n
→
∞
n
|
a
n
|
2
1
/
n
=
lim inf
n
→
∞
(
n
|
a
n
|
)
2
=
0
{\displaystyle \liminf _{n\to \infty }{\frac {n|a_{n}|^{2}}{1/n}}=\liminf _{n\to \infty }(n|a_{n}|)^{2}=0}
, that is,
lim inf
n
→
∞
n
|
a
n
|
=
0
{\displaystyle \liminf _{n\to \infty }n|a_{n}|=0}
.
See also
Convergence tests
Direct comparison test
References
Further reading
Rinaldo B. Schinazi: From Calculus to Analysis. Springer, 2011, ISBN 9780817682897, pp. 50
Michele Longo and Vincenzo Valori: The Comparison Test: Not Just for Nonnegative Series. Mathematics Magazine, Vol. 79, No. 3 (Jun., 2006), pp. 205–210 (JSTOR)
J. Marshall Ash: The Limit Comparison Test Needs Positivity. Mathematics Magazine, Vol. 85, No. 5 (December 2012), pp. 374–375 (JSTOR)
External links
Pauls Online Notes on Comparison Test
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