Calculus II
Contents
Series
Integrals
Definite integrals
Some primitives
Numerical methods
Exercices
© The scientific sentence. 2010
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Calculus II: The criteria of convergence of series
It is often difficult to find the general term
of a series. Often, we need just to know whether
a series is convergent or divergent. In this
case we use the criteria.
1. The divergence criterion
1.1. The rules are:
a) Rule 1:
If the series Σ1∞ an converges, then the related sequence {an} converges,
that is :
lim an = 0
n → + ∞
b) Rule 2:
The contrapositive (converse) proposition is of course true.
If the sequence {an} diverges, then the associated series Σ1∞ an diverges.
Or
If
lim an ≠ 0
n → + ∞
Then the related series Σ1∞ an diverges.
c) Rule 3:
But the reciprocal proposition is not true. That is:
If the sequence {an} converges, the associated series Σ1∞ an is not convergent.
Or
If
lim an = 0
n → + ∞
the associated series Σ1∞ an is not convergent.
1.2. Examples:
b) Rule 1:
The series Σ1∞ (n/(n + 1)) converges,
then the related sequence {1/n(n + 1)} converges, that is :
lim (1/n(n + 1)) = 0
n → + ∞
b) Rule 2:
lim (n(n + 1)/2) ≠ 0
n → + ∞
Then the related series Σ1∞
(n(n + 1)/2) diverges.
c) Rule 3:
To be convergent, the codition an = 0 is necessary for the series
Σ1∞ an, but not sufficient. That is the proposition:
(Σ1∞ an converges) implies
(lim an = 0
n → +∞),
but the converse is not true:
(lim an = 0
n → +∞)
does not imply (Σ1∞ an converges).
Here is an example, the harmonic series:
(lim (1/n) = 0
n → +∞)
and yet
(Σ1∞(1/n) diverges.
2. D'Alembert criterion
The d'Alembert criterion or the ratio criterion is a test
to define the convergence of a series.
The ratio test is written as:
Let the series: Σ1∞ an
L = lim |an+1/an|
n → + ∞
The d'Alembert's ratio states:
if L < 1 then the series converges,
if L > 1 then the series does not converge,
if L = 1 or the limit does not exists,
then the test is not conclusive.
There are convergent series for which L = 1 and
divergent series for which L = 1. Therefore, when
L = 1 or L fails to exist, test is inconclusive.
The d'Alembert's ratio test is efficient when the series
for which the terms contain exponents "n" or n!.
Examples
a)
Σ1∞
(n2n/n!)
| L = | lim
n → + ∞ | |an+1/an|
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an+1/an =
(n+1) n! 2n+1/n (n + 1)!2n = 2/n
|an+1/an| = |2/n| = 2/n
L < 1, the series converges.
b)
Σ1∞
((- 1)n en2)
| L = | lim
n → + ∞ | |an+1/an|
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an+1/an =
(- 1)n+1 e(n+1)2)/en2) =
(- 1) e(2n+1)
|an+1/an| =
|(- 1) e(2n+1)| = e(2n+1)
| L = | lim
n → + ∞ | e(2n+1) = + ∞
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The series diverges.
c)
The harmonic series Σ1∞
(1/n) diverges. The d'Alembert test is not efficient
to conclude for this series. Indeed:
| L = | lim
n → + ∞ | |an+1/an|
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an+1/an =
(1/(n + 1))/(1/n) = n/(n + 1)
|an+1/an| = n/(n + 1)
| L = | lim
n → + ∞ | n/(n + 1) = n/n = 1
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We cannot conclude.
3. Exercises
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