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1.4. Tests for Convergence of Positive Terms Series

Comparison tests for positive terms series make it possible to define whether a number series is convergent (divergent) or not by comparing it with another series that is known as convergent (divergent).

Theorem 7	(Comparison test for positive terms series). Let
			∞
		a1 + a2 + ... + an + ... = ∑ an ,			(1.8)
			n=1
			∞
		b1 + b2 + ... + bn + ... = ∑ bn ,			(1.9)
			n=1
be two series of positive terms, so that
		0 ≤ an ≤ bn			(1.10)
		∞	∞		∞
for all n. Then, if		∑ bn converges,	∑ an converges as well; and if		∑ an di-
∞		n=1	n=1		n=1
∞
verges, then ∑ bn		diverges as well.
n=1
Proof. We form the partial sums of (1.8) and (1.9)
		Sn = a1 + a2 + ... + an , σn = b1 + b2 + ...+ bn .
It follows from (1.10) that Sn ≤ σn			for all n = 1, 2, ....	lim σn	= σ of its
(1) Suppose that series (1.9) converges, i. e, there exists				lim σn	= σ of its
				n→∞
partial sum. Since the terms of these series are positive, then				0 < σn < σ, and it

follows by (1.10) that 0 < Sn < σ for n = 1, 2, ... .

Thus, all the partial sums Sn of (1.8) are bounded and increase with n, since an > 0 for all n. Consequently, the sequence of partial sums {Sn } is convergent,

i. e., there exists lim Sn = S, which implies that		∞
		∑ an is a convergent series.
n→∞		n=1
Now, from the inequality 0 < Sn < σ, which holds for all natural n, we ob-
tain the inequality	0 ≤ S ≤ σ if n → ∞, i. e., the sum S of (1.8) does not exceed
the sum σ of the convergent series (1.9).
	∞
(2) Suppose that ∑an diverges. Since all an > 0, then Sn increases with n,
and hence lim Sn	n=1	σn ≥ Sn (n = 1, 2, ...) we get
	= + ∞. From the inequality
n→∞
lim σn = + ∞, i. e.,	∞
	∑ bn diverges.
n→∞	n=1

Remark 1. Theorem 7 is valid even when (1.10) holds not for all n, but only

begins with some k, i.e., for all n ≥ k, since when we drop a finite number of terms in the beginning, we do not violate the convergence of the series.

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Remark 2. Theorem 7 is also valid for a more general inequality an ≤ λbn (λan ≤ bn ) (n = 1, 2, ...), where (λ > 0).

Theorem 8 (Comparison test for positive terms series). If there exists a finite nonzero

lim	an	= L	(0 < L < +∞),

n→∞ b
	n

then (1.8) and (1.9) converge or diverge simultaneously.

Proof. Indeed, from the existence of the above limit it follows that for any ε > 0 is subjected to the condition L — ε > 0, there exists a number N such that

for all n > N

				an		− L	< ε.
						− L	< ε.
				b
	an				n
or L − ε <	an	< L + ε.
or L − ε <		< L + ε.
	b
	n		< (L + ε)bn		for n > N.
Hence (L − ε)bn < an			< (L + ε)bn		for n > N.
					∞
If (1.9) converges,			so does		∑ (L + ε)bn . But since an < (L + ε)bn			for
					n=1
n > N, then, by Theorem 7, series (1.8) will converge as well.
			∞					< an
If (1.9) diverges then			∑ (L − ε)bn diverges as well. And since (L − ε)bn					< an
			n=1

for n > N, then, by Theorem 7, series (1.8) also diverges.

D’Alembert’s test for convergence of a series is given by the following theorem.

∞

Theorem 9 (D’Alembert’s test). Consider ∑ an , where an > 0. If there exists

n=1

lim an+1 = l,

n→∞ an

then for 0 ≤ l < 1 the series converges, and for l > 1 the series diverges.

Proof. Suppose that there exists the finite limit lim	an+1	= l, where
Proof. Suppose that there exists the finite limit lim	a	= l, where
n→∞	a
	n

l < 1. Take any number q so that l < q < 1. Then, for any ε > 0, e. g., for ε = q – l, there exists a number N such that for all n ≥ N we will have

an + 1 −l < q – l. an

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Specifically, we will have
	an+1	−l < q – l or	an+1	< q.
	a	−l < q – l or	a	< q.
	a		a
	n		n

Hence an+1 < an q for all n ≥ N. From the latter inequality we will obtain, if n assumes the values N, N + 1, N + 2, ...

aN +1 < aN q,

aN + 2 < aN +1q < aN q2 , aN + 3 < aN + 2q < aN q3 ,

.................................

So, the terms of the series

aN +1 + aN + 2 + aN + 3 + ...

are not larger than the corresponding terms of the series aN q + aN q2 + aN q3 + ...,

which converges as a series of the terms of a geometric progression with the ratio

q, where 0 < q < 1. According to comparison test, the series										aN +1 + aN + 2 +
							∞
+aN + 3 + ... converges, and hence ∑ an converges too.
							n=1
If l > 1, then beginning with a certain number N we will have
				aN+1		>1 or aN +1 > aN > 0.
						>1 or aN +1 > aN > 0.
				aN
Consequently, lim an ≠ 0								∞	diverges, since the required
Consequently, lim an ≠ 0					and the series ∑ an				diverges, since the required
	n→∞							n=1
test for convergence is not satisfied.								n=1
test for convergence is not satisfied.
Remark. If lim		an+1	= 1,		then the D’Alembert’s test gives no answer as to
Remark. If lim			= 1,		then the D’Alembert’s test gives no answer as to
	n→∞ a
		n
whether a series converges or not.
									∞
Theorem 10	(Cauchy’s test). Consider the series								∑ an , an > 0,	n = 1, 2, ... . If
	there exists a finite								n=1
	there exists a finite

							lim n a = λ,
							n→∞	n
							n→∞

then for 0 ≤ λ < 1 the series converges, and for λ > 1 diverges.
										13

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Proof. (1) Let λ < 1 . We take q				such that λ < q < 1. Since there exists
lim n a = λ, where λ < q,			then beginning with a certain number N we will
n→∞	n
n→∞				≥ N. And so the terms from aN +1 are
have	n an = q,	whence an	= qn for n	≥ N. And so the terms from aN +1 are
				∞			By the
smaller than the corresponding terms of the convergent series ∑ qn .							By the
				n= N
		∞		∞
comparison test		∑ qn converges, and hence ∑ an converges too.
		n= N		n=1
(2) Let λ > 1.
Then, beginning with a certain N and for all n > N, we will have					n a	n	> 1 or
						n
an > 1. Accordingly, lim an				∞
an > 1. Accordingly, lim an			≠ 0 and the series ∑ an diverges.
		n→∞		n=1
			∞	n=1
			∞
Remark. If λ = 1, then			∑ an can either converge or diverge.
		n=1
				∞
Theorem 11		(Integral test). Let a series ∑ an be positive, i.e. an			≥ 0. Let a
				n=1		≥ 1 so
				n=1
function f(x) be defined, continuous, positive and nonincreasing for x
that f (n) = an . Then the number series				∞
that f (n) = an . Then the number series				∑ an converges, if and only if the inte-
				n=1
	+∞		∞	+∞
gral	∫ f (x)dx converges and ∑an ≤ a1			+ ∫ f (x)dx.
	1		n=1	1

Remark. The theorem also holds for x ≥ a,				where a is any number larger
than unity.
Example. Examine

	∑ 1p	= 1+ 1p	+ 1p + ... + 1p + ...
	∞
	i=1 n	2	3		n

for convergence.

Using the Integral test, we shall investigate the behaviour of this series. We

have f (n) =	1		,	the former satisfying the condition of Integral test. The im-
have f (n) =	n p		,
	n p			1
		∞			dx is known to converge for p > 1 and diverge for p ≤ 1.
proper integral		∫
proper integral		∫		p
		1		x

					∞
Hence,	the			series ∑		1	converges when p > 1 and diverges when
Hence,	the			series ∑		p	converges when p > 1 and diverges when
p ≤ 1.					i=1 n
p ≤ 1.
14

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Specifically,				if p = 1 we	will have the harmonic series	1+	1	+	1	+ ...
Specifically,				if p = 1 we	will have the harmonic series	1+	2	+		+ ...
	1						2	3
		∞
... +		+ ... = ∑	1	, and diverges which has already been shown.
... +	n	+ ... = ∑		, and diverges which has already been shown.
	n	n=1 n		+∞
				+∞
Remark. In the integral ∫					f (x)dx the lower limit may be taken arbitrary,
				1
e.g., m, where m ≥ 1.
				1.5. Alternating Series. Leibniz’ Test
Definition. The number series

					∞
				a1 − a2 + a3 − ... + (−1)n−1 an + ... = ∑ (−1)n+1 an				(1.11)
					n=1

where an > 0 (n = 1, 2, ...) is called the alternating series. For example, the series 1− 12 + 13 − 14 + ... .

is an alternating one.

The following test, known as the Leibniz’s test, holds for alternating series.

Theorem 12		(Leibniz’ test). Suppose that in the alternating series (1.11) all
		an are such that
		an are such that
1)	a1 > a2	> a3 > ... > an > ...
and	lim an	= 0.
2)	lim an	= 0.
	n→∞

Then, the series converges, its sum S is positive and does not exceed the first

term, i.e., 0 < S ≤ a1.

Proof. We take the even partial sum S2n and write it as

	S2n = a1 − a2 + a3 − a4 + ... + a2n−1 − a2n =
	= (a1 − a2 ) + (a3 − a4 ) + ... + (a2n−1 − a2n ).
where each difference is positive (from the given condition 1). It follows that
	S2n ≤ a1 − (a2 − a3 ) − (a4 − a5 ) − ... − (a2n− 2 − a2n−1 ) − a2n ≤ a1.
The	sequence {S2n } thus increases monotonically and is bounded i.e.,
0 < S2n	≤ a1 for all n.
	15

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Consequently, it has the limit	lim S2n	= S so that 0 < S ≤ a1.
	n→∞
The odd partial sum S2n+1 will be
S2n+1 = S2n + a2n+1		(n = 1, 2, ...).
We have proved that lim S2n	= S, i. e., the series converges. In particular,
n→∞

from the inequality 0 < S ≤ a1 it follows that the sum of the series is positive. Remark. The theorem is valid if the condition at which {S2n } is monotonic

is met starting with a certain N for all n ≥ N, so that the discarding of a finite number of terms does not change the convergence of the series.

1.6. Absolute and Conditional Convergence

Let’s consider the series

a1 + a2 + a3 + ... + an + ...

and prove the following theorem:

∞

Theorem 13

If the series

∑

converges, then the series ∑ an converges as

well.

n=1

Proof. From the inequality −

≤ an ≤

we get 0 ≤ an +

≤ 2

for

(n = 1, 2, ...).

∞

Let the series ∑

be convergent, then the series

∑ 2

will be conver-

n=1

∞

n=1

gent, and from the comparison test the series

∑ (an +

) will be convergent

∞

n=1

too. But ∑ an

is the difference of two convergent series

n=1

∞

∑

(an +

) – ∑

= ∑ an ,

n=1

therefore it will converge.

∞

Corollary. If ∑

converges, then we have

∑ an

≤ ∑

n=1

∞

When examining ∑

for convergence we can make use of all sufficient

n=1

tests established for series of positive terms.

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∞

Remark. Generally speaking, the convergence of ∑ an does not suggest the

n=1

∞

convergence of ∑ an , i.e. the theorem only gives a sufficient condition for

n=1

∞

convergence of ∑ an .

n=1

∞

Definition. The series of positive and negative terms ∑ an is called abso-

n=1

∞

lutely convergent, if the series ∑ an converges.

n=1

∞

The series ∑ an is called conditionally convergent if it converges and the

n=1

∞

series ∑ an diverges.

n=1

Micromodule 1

EXAMPLES OF PROBLEMS SOLUTION

Example 1. Prove that the series

1 +

+ ...+

... = ∑

∞

1 2

2 3

n(n +1)

n=1

n(n +1)

converges.

Solution. We consider the nth partial sum of the series

Sn = 1

+ ...+

+ ... = ∑

1 .

∞

1 2

2 3

n(n +1)

n=1

n(n +1)

we represent Sn in such form

Using the usual relation

−

n(n+1)

n+1

S =

+ ...+

1 2 2 3 3 4

n(n +1)

= 1

−

+ ...+

−

= 1

−

n +1

Passing to the limit as n → ∞,

we are having

lim Sn

= lim 1

−

= 1.

+ 1

n→∞

By the definition the series converges and its sum is S = 1.

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Example 2. Prove that the series

+ 1

+...+

+... = ∑

∞

1 4

4 7

7 10

(3n− 2)(3n+1)

n=1

(3n− 2)(3n+1)

converges.

Solution. We represent Sn

in the form of

+...+

1 4

4 7

(3n− 2)(3n+1)

1−

−

+...+

−

1−

3n− 2

3n+1

Passing to the limit as n → ∞,

we are having

lim Sn

= lim

−

+ 1

→∞

n→∞

By the definition

the

series

converges

and

its

sum

S =

∑

= 1 .

∞

n=1

(3n− 2)(3n+1)

Example 3. The number series

2 + cos π + ... = ∑ cos π

−1+ 0 +

1 +

∞

n=1

diverges, since

lim a

= lim cos π = cos0 = 1 ≠ 0 (corollary of theorem 6).

n→∞

Example 4. The series

∞

1− 1+ 1−

1+ ... = ∑ (−1)n+1

= lim(−1)n+1

n=1

diverges, since

lim a

does not exist.

n→∞

∞

Example 5. Investigate the number series ∑

for convergence.

n=1

Solution. We have

lim a

= lim

= 0 but the given series diverges.

n→∞

Let’s consider the partial sum

Sn = 1+

+ ... +

= n

= n.

then

Sn >

n .

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Passing to the limit we obtain

lim Sn = ∞.

n→∞

Hence the given series diverges. Example 6. Examine the series

1+ 5 +

7 + ... + 2n + 1 + ... = ∑ 2n + 1

∞

4 5

n + 2

n=1

n + 2

for convergence

2n + 1

Solution. We have

lim a

= lim

= 2 ≠ 0.

n→∞

n + 2

Then the given series is divergent.

∞

Example 7. Examine the series ∑

for convergence.

Solution. We have

n=1 2n +

(n = 0, 1, 2 ...).

≤

∞

Since ∑

converges, then by the comparison test the original series

n=1 2

converges as well.

∞

Example 8. Examine the series ∑

for convergence.

n= 2 ln n

Solution. From the inequality ln n < n follows

for n = 2, 3, ..., and

ln n

∞

the harmonic series

∑

diverges (in this case

∑

diverges too). Then by the

n=1 n

n= 2 n

comparison test the original series diverges.

∞	π
Example 9. Examine the series ∑(1− cos	π	) for convergence.
Example 9. Examine the series ∑(1− cos	n	) for convergence.
n=1	2

Solution. Using the inequality sin x < x, which holds for all x ≥ 0, we find

that

0 < 1− cos

2sin

≤ 2(

)

π2

2 2n

∞

for n = 1, 2, ... . Here λ =

and

∑

— converges. By comparison test

n=1

(and considering Remark 2), the original series converges.

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∞	π	for convergence.
Example 10. Examine the series ∑ sin	π	for convergence.
n=1	n

∞ 1

Solution. We compare the given series with the harmonic series ∑ and have

n=1 n

			π
lim	an	= lim	sin n	= π 1 = π ≠ 0.
			1
n→∞ bn		n→∞

The harmonic series diverges, therefore, by the corollary, the original series diverges too.

∞

Example 11. Examine the series ∑

for convergence.

− n

n=1

∞

Solution. We take for comparison the convergent series

∑

(theorem 8).

Then

n=1

lim

= lim

2n − n

= lim

= 1 ≠

n→∞ bn

n→∞

1−

since lim

= 0. Therefore, the original series converges.

n→∞ 2n

∞

Example 12. Examine the series ∑

for convergence.

n=1

(n + 1)2

Solution. We have a

and a

n+1

2n+1

Then

(n + 1)2 2n

)2 =

lim

n+1

= lim

(1+

< 1.

2n+1 n2

n→∞

n→∞ 2

By the D’Alembert’s test, the series converges.

∞

Example 13. Examine the series ∑

for convergence.

n=1

(n + 1)n+1

(n + 1)n

Solution. We have

and a

;

n+1

(n + 1)!

(n + 1)n n!

)n = e > 1.

lim

= lim

= lim(1+

n!nn

n→∞

By the D’Alembert’s test, the series diverges.

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