44- Alternating Series; Conditional convergence VU

Lecture # 44

Alternating Series; Conditional convergence

• So far we have seen infinite series that have positive terms only.
• We have also defined the Limit and the Sum of such infinite series
• Now we look at series that have terms which have alternating signs, known as
Alternating series

Example

1) 1-1+1-1+…
2) 1+2-3+4….

More generally, an alternating series has one of the two following forms
∞

∑ (−1)
k =1
k +1
ak = a1 − a2 + a3 − a4 + ... (1)
∞

∑ (−1) a
k =1
k
k =−a1 + a2 − a3 + a4 − ... (2)

Note that all the terms ak are to be taken as being positive.

The following theorem is the key result on convergence of alternating series.

Theorem11.7.1
(Alternating Series Test)
An alternating series of either form (1) or (2) converges if the following two conditions are satisfied:

(a) a 1 >a 2 >a 3 ……..>a k >……..

(b) lim ak = 0
k →+∞
This theorem tells us when an alternating series converges.

Example

Use the alternating series test to show that the following series converge
∞
1
∑ (−1)
k =1
k +1

Solution:
1 1
The two conditions in the alternating series test are satisfied since ak => =
ak +1
k k +1
1
=
and lim ak = 0
lim
k →+∞ k →+∞ k

Note that this series is the Harmonic series, but alternates. It’s called the Alternating Harmonic
Series.
The harmonic series diverges. The alternating Harmonic series converges.
Now we will look at the errors involved in approximating an alternating series with a partial sum.

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44- Alternating Series; Conditional convergence VU

Theorem 11.7.2
If an alternating series satisfies the conditions of the alternating series test, and if the sum S of the
series is approximated by the nth partial sum S n , thereby resulting in an error of S- S n ,then
| S- S n | < a n+1
Moreover, the sign of error is the same as that of the coefficient of a n+1 in the series.

Example
The alternating series
1 – 1/2 + 1/3 – 1/4 +…….+ (-1)k+1 1/k +…….
satisfies the condition of the alternating series test ; hence the series has a sum S, which we know
must lie between any two successive partial sums. In particular, it must lie between

S 7 = 1 – 1/2 + 1/3 – 1/4 + 1/5 – 1/6 + 1/7 = 319/420

And

S 7 = 1 – 1/2 + 1/3 – 1/4 + 1/5 – 1/6 + 1/7 – 1/8 = 533/840

So

533/840 < S < 319/420

If we take S = ln2 then

533/840 < ln2 < 319/420

0.6345 < ln2 < 0.7596

The value of ln2, rounded to four decimal places, is .6931, which is consistent with these inequalities.
It follows from Theorem 11.7.2 that

| ln2 – S 7 | = | ln2 – 319/420 | < a 8 =1/8

And

| ln2 – S 8 | = | ln2 – 533/840 | < a 9 =1/9

Absolute and Conditional Convergence

The series
1 – 1/2 - 1/22 +1/23 + 1/24 – 1/25 – 1/26 +………….

does not fit in any of the categories studied so far, it has mixed signs , but is not alternating. We shall
now develop some convergence tests that can be applied to such series.

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Definition 11.7.3
∞
A series ∑u
k =1
k = u1 + u2 + ........... + uk + .......

is said to converge absolutely, if the series of absolutes values

∞

∑u
k =1
k = u1 + u2 + ........... + uk + ....... converges.
Example
The series
1 - 1/2 - 1/22 +1/23 + 1/24 - 1/25 - 1/26 +………….
Converges absolutely since the series of absolute values
1 + 1/2 + 1/22 +1/23 + 1/24 + 1/25 + 1/26 +………
is a convergent geometric series.
On the other hand, the alternating harmonic series
1 - 1/2 + 1/3 - 1/4 + 1/5 - ………….
does not converge absolutely since the series of absolute values
1 + 1/2 + 1/3 + 1/4 + 1/5 + ………….
diverges.
Absolute convergence is of importance because of the following theorem.

Theorem 11.7.4
∞
If the series ∑u
k =1
k = u1 + u2 + ........... + uk + .......

∞
Converges, then so does the series ∑u
k =1
k = u1 + u2 + ........... + uk + .......

In other words, if a series converges absolutely, then it converges.

Since the series

1 - 1/2 - 1/22 +1/23 + 1/24 - 1/25 - 1/26 +………….
Converges absolutely. It follows from Theorem 11.7.4 that the given series converges.

Example

∞
cos k
Show that the series ∑
k =1 k
2
converges.

Solution: Since cos k ≤ 1 for all k,

cos k 1
2
≤ 2
k k

thus

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∞ ∞
cos k cos k
∑
k =1 k
2 ∑
k =1 k
2

converges by the comparison test, and consequently

∞
cos k
∑
k =1 k
2
converges.

If ∑ uk diverges, no conclusion can be drawn about the convergence or

divergence of ∑uk
For example, consider the two series
1 - 1/2 + 1/3 - 1/4 +…….+ (-1)k+1 1/k +……. (A)
- 1 - 1/2 - 1/3 - 1/4 -…….- 1/k -……. (B)
Series (A), the alternating harmonic series, converges, whereas series (B), being a constant times the
harmonic series, diverges.

Yet in each case the series of absolute values is

1 + 1/2 + 1/3 +…… + 1/k + ……..
which diverges. A series such as (A), which is convergent, but not absolutely convergent, is called
conditionally convergent.

Theorem 11.7.5
(Ratio Test for Absolute Convergence)
uk +1
Let ∑u k be a series with nonzero terms and suppose that ρ = lim
k →∞ uk
(a) If ρ <1, the series ∑u k converges absolutely and therefore converges.

(b) If ρ >1 or ρ = +∞ , then the series ∑u k diverges.

(c) If ρ =1, no conclusion about convergence or absolute convergence can be drawn from this
test.

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Power Series in x

If c 0 , c 1 , c 2 , ………. are constants and x is a variable, then a series of the form

∞

∑c x
k =o
k
k
=c0 + c1 x + c2 x 2 + ........... + ck x k + ....... is called a power series in x.

Some examples of power series in x are

∞

∑x
k =0
k
=1 + x + x 2 + x 3 + ..........
∞
xk x 2 x3
∑
k =0 k !
=1 + x + + + .........
2! 3!
∞
x2k x2 x4 x6
∑ (−1)k
k =0 (2k )!
=−
1 + − + ........
2! 4! 6!

Theorem 11.8.1
For any power series in x, exactly one of the following is true:
(a) The series converges only for x=0
(b) The series converges absolutely (and hence converges) for all real values of x.
(c) The series converges absolutely (and hence converges) for all x in some finite open interval
(-R, R), and diverges if x<-R or x>R. At either of the points x = R or x = -R, the series may
converge absolutely, converge conditionally, or diverge, depending on the particular series.

Radius and Interval of Convergence

Theorem 11.8.1 states that the set of values for which a power series in x converges is always an
interval centered at 0; we call this the interval of convergence, corresponding to this interval series
has radius called radius of convergence.

Example:
Find the interval of convergence and radius of convergence of the following power series.

∑
k =1
xk

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44- Alternating Series; Conditional convergence VU

Solution: We shall apply the ratio test for absolute convergence. We have

uk +1 x k +1
=ρ lim = lim = =
lim x x
k →∞ u k →∞ x k k →∞
k

So the ratio test for absolute convergence implies that the series converges absolutely if ρ= x < 1
and diverges if ρ= x > 1 .The test is inconclusive if |x| =1 ( i.e. x = 1 or x = -1 ), so
convergence at these points must be investigated separately . At these points the series becomes

∑1
k =0
k
= 1 + 1 + 1 + 1..... x=1
∞

∑ ( −1) = 1 − 1 + 1 − 1 + ......
k
x=-1
k =0
Both of which diverge; thus, the interval of convergence for the given power series is
(-1,1), and the radius of convergence is R = 1.

Power series in x-a

∞

∑ c ( x − a)
k =0
k
k
= c0 + c1 ( x − a ) + c2 ( x − a ) 2 + ......... + ck ( x − a ) k + .......
This series is called power series in x-a. Some examples are
( x − 1)
k
∞
( x − 1) ( x − 1) 2 ( x − 1)3
∑
k =0 k + 1
= 1 +
2
+
3
+
4
+ .......... a=1

∞
(−1) k ( x + 3) ( x + 3) 2 ( x + 3)3
∑
k =0 k!
= 1 − ( x + 3) +
2!
−
3!
+ ......... a=-3

Theorem 11.8.1
For any power series in ∑
ck ( x − a ) k , exactly one of the following is true:
(a) The series converges only for x=a
(b) The series converges absolutely (and hence converges) for all real values of x.
(c) The series converges absolutely (and hence converges) for all x in some finite open interval
(a-R,a+R), and diverges if x<a-R or x>a+R. At either of the points x =a-R or x =a+R, the
series may converge absolutely, converge conditionally, or diverge, depending on the
particular series.

It follows from this theorem that now interval of convergence is centered at x=a.

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