6 Series - Pennsylvania State University

[Pages:14]6 Series

We call a normed space (X, ? ) a Banach space provided that every Cauchy

sequence (xn) in X converges. For example, R with the norm ? = |?| is an example of Banach space. Now let (xn) be a sequence in X. Define a new sequence (sn) in X by

n

sn = xk, n N.

k=1

The sequence (sn) is called a series in X and is written as xk or k xk. The n-th term sn of the sequence (sn) is called the nth partial sum and xk

is called the kth summand of the series k xk.

Definition 6.1. The series k xk converges or is convergent if the sequence

(sn) of partial sums converges. The limit limn sn is called the value of

the series

k xk and is written as

k=1

xk

Finally,

k xk diverges or is

divergent if the sequence (sn) diverges in X.

Proposition 6.2 (Necessary condition for convergence). If the series xk converges, then xn 0.

Proof. Denoting by sn the nth partial sum of the series xk, we have sn x for some x. Since

xn = sn+1 - sn x - x = 0,

the result follows.

Example 6.3. The series

(1/k!) converges in R. Its value is

k=0

1/k!

=

e.

Example 6.4. The series

1/k2 converges in R. Indeed, if sn =

n k=1

1/k2

,

then

sn =

n

1/k2 1 +

n

1 k(k -

1)

=

1

+

n

k

1 -

1

-

1 k

=

1

+

1

-

1 n

<

2.

k=1

k=2

k=2

So, the sequence (sn) of partial sums is bounded in R and since it is also increasing, it converges.

Example 6.5. The harmonic series 1/k diverges. Indeed, if sn =

then

|s2n

-

sn|

=

n

1 +

1

+

...

+

1 2n

n 2n

=

1 2

n k=1

1/k,

showing that (sn) is not a Cauchy sequence and this implies that k 1/k is dirvergent.

30

Example 6.6. The geometric series ak, where a R satisfies |a| < 1,

converges. Indeed,

sn =

n

ak

=

ak+1 - a-1

1

k=0

and since |a| < 1, we have ak+1 0 showing that

lim

n

sn

=

ak = 1 . 1-a

k=0

If |a| 1, then k ak diverges.

Proposition 6.7. Let ak and bk be convergent series in a Banach space X and let R. Then

? The series

k=1

bk

.

? The series

(ak + bk) converges and

k=1(ak + bk) =

k=1

ak

+

k=1(ak) converges and

k=1(ak) =

k=1

ak

.

Theorem 6.8 (Cauchy criterion). For a series xk in a Banach space (X, ? ) the following conditions are equaivalent.

(a) The series xk converges.

(b) For every > 0 there is N N such that

m

xk <

k=n+1

for all m > n N .

Proof. Let sn =

n k=1

xk

be

the

nth

partial

sum.

Then sm-sn

=

m k=n+1

xk

if m > n.Thus (sn) is Cauchy in X if and only if the condition (b) holds

true. By assumption, (X, ? ) is a Banach space and the result follows.

Theorem 6.9. If ak is a series in R and ak 0 for all k N, then ak converges if and only if the sequence (sn) of partial sums is bounded. In this case, the series has the value equal to supnN sn.

Proof. Since ak 0 for all k N, the sequence (sn) of partial sums is increasing. Hence (sn) converges if and only if (sn) is bounded.

31

6.1 Alternating series

Definition 6.10. A series ak in R is called alternating if ak and ak+1 have opposite signs. Any such series can be written as ? (-1)kak with ak 0 for all k N.

Theorem 6.11 (Leibniz criterion). Let (ak) be a decreasing null sequence such that ak 0 for all k N. Then the alternating series (-1)kak converges in R.

Proof. Since

s2n+2 - s2n = -a2n+1 + a2n+2 0,

the sequence (s2n) is decreasing. Similarly,

s2n+3 - s2n+1 = a2n+2 - a2n+3 0,

implying that (s2n+1) is increasing. Moreover, s2n+1 s2n so that s2n+1 s2 and s2n 0 for all n 1. Hence, there are s and t such that s2n s and s2n+1 t. Also

t

-

s

=

nlim(s2n+1

-

s2n)

=

lim

n

a2n+1

=

0

implying that t = s. Take > 0. Then there are N1, N2 N such that

|s2n - s| < , 2n N1 and |s2n+1 - s| < , 2n + 1 N2.

Thus, |sn - s| < for all n > 2 max{N1, N2}.

Example 6.12. The assumption that (an) is decreasing is necessary. For

example, set

a2n

=

1 2n

and

a2n-1

=

(2n

1 -

1)2 .

Then

a2n-1

=

1 (2n-1)2

1 2n-1

so

that

0

an

1 n

showing

that

an

0.

Since a2n-1 < a2n for n 2, the sequence (an) is not decreasing. However,

the series (-1)nan diverges. Arguing by contradiction we assume that the

sequence (sn) of partial sums converges. We have

2n

n

n

s2n = (-1)kak = (-1)a2k-1 + a2k

k=1

k=1

k=1

=-

n

(2k

1 -

1)2

+

n

1 2k

k=1

k=1

= -An + Bn,

32

where An = -

n

1

k=1 (2k-1)2

and

Bn

=

n1 k=1 2k

=

1 2

n k=1

1 k

.

Since

the

series

1 k1 (2k-1)2

converges,

the sequence (An) converges.

Consequently,

the sequence Bn = sn + An converges. But this contradicts the fact that the

series

k1

1 k

diverges.

6.2 Absolute convergence

Definition 6.13. The series xk in a Banach space (X, ? ) converges absolutely or is absolutely convergent if the series xk is convergent in R. The series xk is conditionally convergent if xk converges but xk diverges.

Example 6.14. The alternating harmonic series (-1)k/k converges in

view of Leibniz criterion, however, the series of absolute values 1/k diverges. Hence the series (-1)k/k converges conditionally.

Proposition 6.15. Every absolutely convergent series converges.

Proof. The proposition is a consequence of Cauchy criterion and the triangle

inequality. Indeed,

m

m

xk

xk .

k=n+1

k=n+1

6.3 Comparison, Root, and Ratio Tests

From now on we only consider real sequences, however, you are invited to figure out which of the results can be stated in the full generality of Banach spaces.

Proposition 6.16 (Comparison test). Let ak and bn be two sequences of nonnegative terms and assume that

an bn for all n.

Then the following holds true,

(a) If (b) If

bk converges, then so is ak. ak diverges, then so is bk.

33

Example 6.17. Let m 2. Then the series 1/km converges absolutely.

Recall from Example 6.4 that the series

1 k2

converges.

Since

for

m 2,

we have

1 km

1 k2

for all k,

it follows from Proposition 6.16 that the series 1/km is indeed convergent.

Proposition 6.18 (Limit comparison test). Let ak and sequences of positive terms and assume that

an bn

L

(Clearly, L [0, ) or L = ). Then the following holds true,

bn be two

(a) If L (0, ), then the convergence of bk implies the convergence of ak and the divergence of ak implies the divergence of bk.

(b) Assume that L = 0. Then ak converges if bk converges.

(c) Assume that L = . Then ak diverges, if bk diverges.

Proof. We only prove (a). If L (0, ), then there is N so that L/2 = L-L/2 < an/bn < L+L/2 = 3L/2 for n N . So, (L/2)bn < an < (3L/2)bn for n N and the result follows from Proposition 6.16.

Theorem 6.19 (Root test). Let ak be a series of nonnegative real numbers and let

= lim sup n |an|.

Then

? ak converges absolutely if < 1. ? ak diverges if > 1.

If = 1, then both convergence and divergence of ak are possible.

Theorem 6.20 (Ratio test). Let ak be a series of real numbers such that ak = 0 for all k N . Then

? If then

lim

k

ak+1 ak

< 1,

ak converges absolutely.

34

? If then ak diverges.

lim

k

ak+1 ak

> 1,

If limk

ak+1 ak

= 1, then both convergence and divergence of

possible.

ak are

Proposition 6.21 (Cauchy's condensation test). Let (ak) be a sequence

of nonnegative and decreasing terms. Then ak converges if and only if the series 2ka2k converges.

Proof. Denote by sn =

n k=1

and

tn

=

n k=1

2k

a2k

.

Then,

if

n

<

2m,

we

have

sn a1 + (a2 + a3) + . . . + (a2m + . . . + a2m+1-1) a1 + 2a2 + . . . + 2ma2m = tm.

If 2ka2k converges, then the sequence (tm) is bounded and above estimate shows that (sn) is bounded. Hence ak converges. For the converse, note that if n > 2m, then

sn a1 + a2 + (a3 + a4) + . . . + (a2m-1+1 + . . . + a2m )

1 2

a1

+

a2

+

2a4

+

...

+

2m-1a2m

=

1 2

tm.

This shows that if ak converges, then the sequence (tm) is bounded so that 2ka2k converges.

Example 6.22. Consider the series

1 np

with

p

>

0.

sation test implies that

1 np

converges

if

and

only

if

converges. We already know that the geometric series

if

and

only

if

1 2p-1

<

1,

i.e.,

p

>

1.

Then the conden-

2n 2np

1

=

n

2p-1

1n 2p-1

converges

The next theorem is a generalization of the condensation test.

Theorem 6.23 (Schlo?mlich theorem). Let (gk) be a strictly increasing sequence of positive integers such that gk+1 - gk C(gk - gk-1) for some C > 0 and all k N. If (ak) is a decreasing sequence of nonnegative numbers, then

ak converges if and only if (gk+1 - gk)agk converges.

35

Proof. The proof is similar to the proof of the condensation test. Let (Sn) be the sequence of the partial sums of the series ak and (Tn) the sequence of the partial sums of the series (gk+1 - gk)agk . Now, if n < gk, then

Sn Sgk (a1 + . . . + ag1-1) + (ag1 + . . . + ag2-1) + . . . + (agk + . . . + agk+1-1) (a1 + . . . + ag1-1) + (g2 - g1)ag1 + . . . + (gk+1 - gk)agk (a1 + . . . + ag1-1) + Tk.

Consequently, if the sequence (Tk) converges and hence is bounded, then the sequence (Sn) is bounded so that ak converges. Conversely, assume that ak converges. If n > gk, then

CSn CSgk C(ag1+1 + . . . + ag2 ) + . . . + C(agk-1+1 + . . . + agk C(g2 - g1)ag2 + . . . + C(gk - gk-1)agk (g3 - g2)ag2 + . . . + (gk+1 - gk)agk = Tk - (g2 - g1)ag1 .

Consequently, the sequence (Tk) is bounded and the series (gk+1 - gk)agk converges.

Example 6.24. Take gk = 3k. Then the sequence (gk) satisfies the assumptions of the above theorem and gk+1 - gk = 3k+1 - 3k = 2 ? 3k. So, the following holds true. Assume that (ak) is a decreasing sequence of nonnegative numbers, then

ak converges if and only if 3ka3k converges.

Na3lnnod13w-3c1lnco13onnns13c-ie1f1,atnhnitedcofsoonenlvqlloeuyrwegnisefcset.h(aH33tll3ne1nn3n3nnlcn)e3=i-s1

decreasing

= 3n

3n ln 3

> 1 and

1 3ln n

con-

verges.

Consider gk = k2, then the sequence (gk) satisfies the assumptions of the

theorem. Then the following is true. If (ak) is a decreasing sequence of

nonnegative numbers, then

ak converges if and only if kak2 converges.

Indeed, since (k + 1)2 - k2) = 2k + 1, it follows from Theorem ?? that ak converges if and only if (2k + 1)ak2 converges. Since kak2 (2k + 1)ak2

36

3kak2, the comparison test shows that (2k + 1)ak2 converges if and only if

cthonekvaseekrq2guecseonnicfveearn1g/des2.onsnhl)yowiisfindtgheceoruesraesrciilneasgima.nN2donncw2onc=voenrsgidese2nrnttohcoe0n,sveterhriegesess.eri2Ue1ssnin. gSi2nth1cnee

ratio test, we see that

n 2n

converges

and

so

1 2n

converges.

6.4 The Dirichlet and Abel Test

We start with the following lemma.

Lemma 6.25 (Abels' lemma). Let (xn) and (yn) be two sequences of real numbers. Let (sn) be a sequence of partial sums of (yn) and s0 = 0. Then

m

m-1

xkyk = (xmsm - xn+1sn) +

(xk - xk+1)sk.

k=n+1

k=n+1

Proof. Note that yk = sk - sk-1. So,

m

m

xkyk =

xk(sk - sk-1)

k=n+1

k=n+1

= xn+1(sn+1 - sn) + xn+2(sn+2 - sn+1) + . . . + xm(sm - sm-1)

= xm(xm - xn+1sn) + sn+1(xn+1 - xn+2) + . . . + sm-1(xm-1 - xm)

m-1

= xm(xm - xn+1sn) +

xk(sk - sk-1)

k=n+1

We use this lemma to obtain a test for convergence of the series xkyk.

Theorem 6.26 (Dirichlet's test). Suppose that (xn) is a decreasing sequence such that xn 0 and the sequence (sn) of the partial sums of (yn) is bounded. Then the series xkyk converges.

Proof. Since (sn) is bounded, |sn| C for all n 1. Since (xk) is decreasing, xk - xk+1 0. Thus, by the above lemma,

m

m-1

xkyk |xmsm| + |xn+1sn| +

(xk - xk+1) |sk|

k=n+1

k=n+1

((xn+1 + xm) + (xn+1 - xm))C = 2xn+1C.

37

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