Calculus II
Contents
Series
Integrals
Definite integrals
Some primitives
Numerical methods
Exercices
© The scientific sentence. 2010


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 Σ_{1}^{∞} a_{n} converges, then the related sequence {a_{n}} converges, that is :
lim a_{n} = 0
n → + ∞
b) Rule 2:
The contrapositive (converse) proposition is of course true.
If the sequence {a_{n}} diverges, then the associated series Σ_{1}^{∞} a_{n} diverges.
Or
If
lim a_{n} ≠ 0
n → + ∞
Then the related series Σ_{1}^{∞} a_{n} diverges.
c) Rule 3:
But the reciprocal proposition is not true. That is:
If the sequence {a_{n}} converges, the associated series Σ_{1}^{∞} a_{n} is not convergent.
Or
If
lim a_{n} = 0
n → + ∞
the associated series Σ_{1}^{∞} a_{n} is not convergent.
1.2. Examples:
b) Rule 1:
The series Σ_{1}^{∞} (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 Σ_{1}^{∞}
(n(n + 1)/2) diverges.
c) Rule 3:
To be convergent, the codition a_{n} = 0 is necessary for the series
Σ_{1}^{∞} a_{n}, but not sufficient. That is the proposition:
(Σ_{1}^{∞} a_{n} converges) implies
(lim a_{n} = 0
n → +∞),
but the converse is not true:
(lim a_{n} = 0
n → +∞)
does not imply (Σ_{1}^{∞} a_{n} converges).
Here is an example, the harmonic series:
(lim (1/n) = 0
n → +∞)
and yet
(Σ_{1}^{∞}(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: Σ_{1}^{∞} a_{n}
L = lim a_{n+1}/a_{n}
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)
Σ_{1}^{∞}
(n2^{n}/n!)
L =  lim
n → + ∞  a_{n+1}/a_{n}

a_{n+1}/a_{n} =
(n+1) n! 2^{n+1}/n (n + 1)!2^{n} = 2/n
a_{n+1}/a_{n} = 2/n = 2/n
L < 1, the series converges.
b)
Σ_{1}^{∞}
(( 1)^{n} e^{n}^{2})
L =  lim
n → + ∞  a_{n+1}/a_{n}

a_{n+1}/a_{n} =
( 1)^{n+1} e^{(n+1)}^{2})/e^{n}^{2}) =
( 1) e^{(2n+1)}
a_{n+1}/a_{n} =
( 1) e^{(2n+1)} = e^{(2n+1)}
L =  lim
n → + ∞  e^{(2n+1)} = + ∞

The series diverges.
c)
The harmonic series Σ_{1}^{∞}
(1/n) diverges. The d'Alembert test is not efficient
to conclude for this series. Indeed:
L =  lim
n → + ∞  a_{n+1}/a_{n}

a_{n+1}/a_{n} =
(1/(n + 1))/(1/n) = n/(n + 1)
a_{n+1}/a_{n} = n/(n + 1)
L =  lim
n → + ∞  n/(n + 1) = n/n = 1

We cannot conclude.
3. Exercises

