For D=1, the D’Alembert criterion cannot be used for deciding whether the series is convergent or divergent.. The series is divergent if λ 1the series ∞ n=1a nis convergent, and forR...
Trang 1If D <1, then the series ∞
n=1a n is convergent If D >1, then the series is divergent For
D=1, the D’Alembert criterion cannot be used for deciding whether the series is convergent
or divergent
Example 1 Let us examine convergence of the series∞
n=1n
k x n with x >0, using the D’Alembert criterion.
Taking a n = n k x n, we get
a n+1
a =
1 + 1
n
k
x → x as n → ∞.
Therefore, D = x It follows that the series is convergent for x <1and divergent for x >1.
4 Cauchy’s criterion Suppose that there exists the limit (finite or infinite)
lim
n→∞ n
√
a n = K.
For K <1, the series ∞
n=1a n is convergent; for K >1, the series is divergent For K =1, the Cauchy criterion cannot be used to establish convergence of a series
Remark The Cauchy criterion is stronger than the D’Alembert criterion, but the latter is simpler than the former.
5 Gauss’ criterion Suppose that the ratio of two consecutive terms of a series can be
represented in the form
a n
a n+1 = λ +
μ
n + o
1
n
as n → ∞.
The series ∞
n=1a n is convergent if λ >1or λ =1, μ >1 The series is divergent if λ <1or
λ=1, μ≤ 1
6 Maclaurin–Cauchy integral criterion Let f (x) be a nonnegative nonincreasing
continuous function on the interval1 ≤x<∞ Let f(1 ) = a1, f (2) = a2, , f (n) = a n ,
Then the series ∞
n=1a n is convergent if and only if the improper integral
∞
1 f (x) dx is
convergent
Example 2 The harmonic series ∞
n=1
1
n = 1 + 1
2 +
1
3+· · · is divergent, since the integral
∞
1
1
x dxis divergent In a similar way, one finds that the series
∞
n=1
1
n α is convergent for α >1and divergent for α≤ 1.
8.1.2-2 Other criteria of convergence (divergence) of series with positive terms
1 Raabe criterion Suppose that there exists the limit (finite or infinite)
lim
n→∞ n
a n
a n+1 –1
=R.
Then, forR >1the series ∞
n=1a nis convergent, and forR <1it is divergent ForR =1, the Raabe criterion is inapplicable
Trang 22 Bertrand criterion Suppose that there exists the limit (finite or infinite)
lim
n→∞ ln n
n
a n
a n+1 –1
–1
=B.
Then, forB >1the series ∞
n=1a nis convergent, and forB <1it is divergent ForR =1, the Bertrand criterion is inapplicable
3 Kummer criterion Let b nbe an arbitrary sequence with positive terms Suppose that there exists the limit (finite or infinite)
lim
n→∞
b n a a n n+1 – b n+1
=K.
Then, forK > 0, the series ∞
n=1a n is convergent If K < 0 and the additional condition
∞
n=1
1
b n =∞ holds, then the series is divergent.
Remark. From the Kummer criterion, we obtain the D’Alembert criterion (taking b n = 1), the Raabe
criterion (taking b n = n), and the Bertrand criterion (taking b n = n ln n).
4 Ermakov criterion Let f (x) be a positive monotonically decreasing continuous
function on the interval1 ≤x<∞ Let f(1 ) = a1, f (2) = a2, , f (n) = a n , Then the
following implications hold:
1) If e x f (e x)
f (x) ≤q <1, then the series
∞
n=1
a n is convergent.
2) If e x f (e x)
f (x) ≥ 1, then the series
∞
n=1
a n is divergent.
Here, it suffices to have the inequalities on the left for sufficiently large x≥x0.
5 Generalized Ermakov criterion Let f (x) be the function involved in the Ermakov criterion, and let ϕ(x) be an arbitrary positive monotonically increasing function that has a continuous derivative and satisfies the inequality ϕ(x) > x Then the following implications
hold:
1) If ϕ (x)f ϕ (x)
f (x) ≤q<1, then the series
∞
n=1
a n is convergent.
2) If ϕ (x)f ϕ (x)
f (x) ≥ 1, then the series
∞
n=1
a n is divergent.
Here, it suffices to have the inequalities on the left for sufficiently large x≥x0.
6 Sapogov criterion Let a1, a2, be a monotonically increasing sequence Then the
series
∞
n=1
1– a n
a n+1
as well as
∞
n=1
a n
a n+1 –1
is convergent, provided that the sequence a n is bounded (a n≤L) Otherwise, this series is divergent
Trang 37 Special Cauchy criterion Suppose that a1, a2, is a monotonically decreasing
sequence Then the series ∞
n=1a n is convergent (resp., divergent) if and only if the series
∞
n=12n a2n is convergent (resp., divergent).
8 Abel–Dini criterion If the series ∞
n=1a n is divergent and s ndenotes its partial sum,
then the series ∞
n=1
a n
s n is also divergent, while the series
∞
n=1
a n
s1 +σ
n (σ >0) is convergent
9 Dini criterion If the series ∞
n=1a n is convergent and γ ndenotes its remainder after
the nth term, then the series∞
n=1
a n
γ n–1 is divergent, while the series
∞
n=1
a n
γ σ n–1
(0< σ <1) is convergent
10 Bugaev criterion If the function ϕ (x)u ϕ (x)
is monotone for large enough x,
then the series ∞
n=1u (n) and
∞
n=1ϕ
(n)u ϕ (n)
are convergent or divergent simultaneously
11 Lobachevsky criterion Let u(x) be a monotonically decreasing function defined for all x Then the series ∞
n=1u (n) is convergent or divergent if and only if the series
∞
k=1p k2–k
is convergent or divergent, where p k is defined from the relation u(p k) =2–k.
8.1.3 Convergence Criteria for Arbitrary Numerical Series Absolute
and Conditional Convergence
8.1.3-1 Arbitrary series Leibnitz, Abel, and Dirichlet convergence criteria
1 Leibnitz criterion Suppose that the terms a nof a series∞
n=1a nhave alternating signs,
their absolute values form a nonincreasing sequence, and a n → 0 as n → ∞ Then this
“alternating” series is convergent If S is the sum of the series and s n is its nth partial sum,
then the following inequality holds for the error|S– s n| ≤ |an+1|
Example 1 The series1 – 1
2 2 + 1
3 3 – 1
4 4 + 1
5 5 –· · · is convergent by the Leibnitz criterion Taking
S≈s4= 1 – 1
2 2 + 1
3 3 – 1
4 4, we obtain the error less than a5= 1
5 5 = 0.00032.
2 Abel criterion Consider the series
∞
n=1
a n b n = a1b1+ a2b2+· · · + a n b n+· · · , (8.1.3.1)
where a n and b nare two sequences or real numbers
Series (8.1.3.1) is convergent if the series
∞
n=1
b n = b1+ b2+· · · + b n+· · · (8.1.3.2)
is convergent and the a nform a bounded monotone sequence (|an|< K).
Trang 43 Dirichlet criterion Series (8.1.3.1) is convergent if partial sums of series (8.1.3.2) are bounded uniformly in n,
n
k=1
b k≤M (n =1, 2, ),
and the sequence a n →0is monotone
Example 2 Consider the series ∞
n=1a sin(nx), where a n →0 is a monotonically decreasing sequence.
Taking b n = sin(nx) and using a well-known identity, we find the partial sum
s n=
n
k=1
sin(kx) = cos
1
2x
– cos n+12
x
2 sin 12x (x≠ 2mπ ; m =0, 1, 2, ).
This sum is bounded for x≠ 2mπ:
|s n| ≤ sin11
2x Therefore, by the Dirichlet criterion, the series ∞
n=1a sin(nx) is convergent for any x≠ 2mπ Direct verification
shows that this series is also convergent for x =2mπ(since all its terms at these points are equal to zero).
Remark The Leibnitz and the Able criteria can be deduced from the Dirichlet criterion.
8.1.3-2 Absolute and conditional convergence
1 Absolutely convergent series A series ∞
n=1a n(with terms of arbitrary sign) is called
absolutely convergent if the series ∞
n=1|an|is convergent
Any absolutely convergent series is convergent In order to establish absolute conver-gence of a series, one can use all converconver-gence criteria for series with nonnegative terms
given in Subsection 8.1.2 (in these criteria, a nshould be replaced by|an|)
Example 3 The series1 + 1
2 2 – 1
3 2 – 1
4 2 + 1
5 2 + 1
6 2 –· · · is absolutely convergent, since the series with
the absolute values of its terms, ∞
n=1
1
n2, is convergent (see the second series in Example 2 of Subsection 8.1.2
for α =2).
2 Conditionally convergent series A convergent series ∞
n=1a n is called conditionally
convergent if the series∞
n=1|an|is divergent
Example 4 The series1 – 1
2 + 13 –14 +· · · is conditionally convergent, since it is convergent (by the Leibnitz criterion), but the series with absolute values of its terms is divergent (it is a harmonic series; see Example 2 in Subsection 8.1.2).
Any rearrangement of the terms of an absolutely convergent series (in particular, a convergent series with nonnegative terms) neither violates its absolute convergence nor changes its sum Conditionally convergent series do not possess this property: the terms of
a conditionally convergent series can be rearranged in such order that the sum of the new series becomes equal to any given value; its terms can also be rearranged so as to result in
a divergent series
Trang 58.1.4 Multiplication of Series Some Inequalities
8.1.4-1 Multiplication of series Cauchy, Mertens, and Abel theorems
A product of two infinite series∞
n=0a nand
∞
n=0b nis understood as a series whose terms
have the form a n b m (n, m =0, 1, ) The products a n b mcan be ordered to form a series
in many different ways The following theorems allow us to decide whether it is possible
to multiply series
CAUCHY THEOREM Suppose that the series∞
n=0a nand ∞
n=0b nare absolutely convergent
and their sums are equal to A and B, respectively Then any product of these series
is an absolutely convergent series and its sum is equal to AB The following Cauchy
multiplication formulaholds:
∞
n=0
a n
∞
n=0
b n
=
∞
n=0
n
m=0
a m b n–m
(8.1.4.1) MERTENS THEOREM The Cauchy multiplication formula (8.1.4.1) is also valid if one of the series,∞
n=0a nor∞
n=0b n, is absolutely convergent and the other is (conditionally) conver-gent In this case, the product is a convergent series, possibly, not absolutely converconver-gent ABEL THEOREM Consider two convergent series with sums A and B Suppose that the product of these series in the form of Cauchy (8.1.4.1) is a convergent series with sum C Then C = AB.
8.1.4-2 Inequalities
1 Generalized triangle inequality:
∞
n=1
a n≤∞ n=1
|an|
2 Cauchy inequality (Cauchy–Bunyakovsky inequality):
∞ n=1
a n b n
2
≤∞
n=1
a2
n
∞ n=1
b2
n
3 Minkowski inequality:
∞
n=1
|an + b n|p1
p ≤∞
n=1
|an|p1
p +∞ n=1
|bn|p1
p, p≥ 1
4 H¨older inequality (for p =2coincides with the Cauchy inequality):
∞
n=1
a n b n≤∞
n=1
|an|p1
p∞ n=1
|bn|p– p1p–1
p , p>1
5 An inequality with π:
∞
n=1
a n
4
≤π2∞
n=1
a2
n
∞
n=1
n2a2
n
In all these inequalities it is assumed that the series in the right-hand sides are convergent
Trang 68.1.5 Summation Methods Convergence Acceleration
8.1.5-1 Some simple methods for calculating the sum of a series
THEOREM1 Suppose that the terms of a series ∞
n=1a n can be represented in the form
a n = b n – b n+1, where b n is a sequence with a finite limit b ∞ Then
∞
n=1
a n = b1– b ∞.
Example 1 Let us find the sum of the infinite series S =∞
n=1
1
n (n +1).
We have
a = 1
n (n + a) =
1
n – 1
n+ 1 =⇒ b n=
1
n.
Since b1= 1 and lim
n→∞ b n= 0, we get S = 1.
THEOREM2 Suppose that the terms of a series ∞
n=0a ncan be represented in the form
a n = b n – b n+m, where m is a positive integer and the sequence b n has a finite limit b ∞ Then
∞
n=1
a n = b0+ b1+· · · + b m–1– mb ∞
THEOREM3 Suppose that the terms of a series ∞
n=0a ncan be represented in the form
a n = α1b n+1+ α2b n+2+· · · + α m b n+m, (8.1.5.1)
where m≥ 2is a fixed positive integer, the sequence b n has a finite limit b ∞ , and α ksatisfy the condition
α1+ α2+· · · + α m =0 (8.1.5.2) Then the series is convergent and
∞
n=0
a n = α1b1+(α1+α2)b2+· · ·+(α1 +α2+· · ·+α m–1)b m–1+[α2+2α3+· · ·+(m–1 )α m ]b ∞
(8.1.5.3)
Example 2 For the series ∞
n=0
4n+ 6
(n +1)(n + 2)(n + 3), we have
a = 4n+ 6
(n +1)(n + 2)(n + 3) =
1
n+ 1 +
2
n+ 2–
3
n+ 3, which corresponds to the following values in (8.1.5.1):
α1 = 1, α2 = 2, α3 = –3, b n= 1
n Thus, condition (8.1.5.2) holds: 1 + 2 – 3 = 0, and the sequence bn tends to zero: b ∞= 0 Using (8.1.5.3), we find the sum of the series
∞
n=0
4n+ 6
(n +1)(n + 2)(n + 3) =1 ⋅ 1+ (1+2)21 = 52.
Trang 78.1.5-2 Summation of series with the help of Laplace transforms.
Let a(k) be the Laplace transform of a given function f (x),
a (k) =
∞
0 e
–kx f (x) dx.
Then the following summation formulas hold:
∞
k=1
a (k) =
∞
0
f (x) dx
e x–1 ,
∞
k=1
(–1)k a (k) = –
∞
0
f (x) dx
e x+1 ,
(8.1.5.4)
provided that the series are convergent
Example 3 It is easy to check that
a
k2+ a2 =
∞
0
e–kx sin(ax) dx.
Therefore, using the first formula in (8.1.5.4), we get
∞
k=1
1
k2+ a2 =
1
a
∞
0
sin(ax) dx
e – 1 =
π
2a coth(πa) – 1
2a2.
8.1.5-3 Kummer and Abel transformations Acceleration of convergence of series
1◦ Kummer transformation Consider a series with positive (nonnegative) terms
∞
n=1
and an auxiliary series with a finite sum
B =
∞
n=1
Suppose that there is a finite limit
K= lim
n→∞
b n
a n ≠ 0 Under these conditions, the series (8.1.5.5) is convergent and the following identity holds:
∞
n=1
a n = K B +
∞
n=1
1– 1
K
b n
a n
The right-hand side of (8.1.5.7) is called the Kummer transformation of the given series