- 1 -
(b) (2.1)
(c)
In general, an infinite series means an expression of the form
x
2
3
4
x x
2
x
3 ...
4
2
2
2
4 2
1 3
2
4
2 ...
2
(a) 1
2 2
2 3
2 4
2 ...
There are many infinite series besides geometric series. Here are some examples:
a
1 a
2 a
3 a
4 ... a
n ...
Where
formula or rule. The three dots mean that the series never ends.
(b)
(c) (2.2)
a
n(one for each positive integer n) are numbers or functions given by some
3
n 1 nx x
2
2
x ( 1) x
... ...
2
3
4
n 1 n
x 2
x 3 x
4
x ( 1) x
... ...
n
2
3
4
n 2
1 2
2 2
3 2
4 n
... ...
2
(a) 1
2 2
2 3
2 4
2... n
2 ...
n 1
1
22
23
24
2... n
2We can write the series in a shorter abbreviated form using a summation sign
followed by the formula for the nth term. For example (2.1a) would be written
n 1
3 3 n 1 n
2
n 1 ! ( 1) x . . .
6 x 2 x x x 1. Definitions and Notation
A series is an infinite ordered set of terms combined together by the addition operator. The term infinite series is used to confirm the fact that series contain an infinite number of terms.
(d) n 1 !
Chapter Two : Infinite Series, Power Series
The series (2.2d) would be written
a) 2, 4, 8, 16, 32, ….
2 4, , 9 27
8 18
b)
, , ...81
3 27 ....
81
n
nS 1 r
a 1 r
4 3
2 ...
9 3
2
2 1
3 3
2
1 3
n
2 2 1
3
n
2
n
n
as n increases,
n
3 2
decreases and approaches zero. Then the sum of n terms 2. The Geometric Series
approaches 2 as n increases, and we say that the sum of the series is 2.
1 r a If we have some bacteria in the Lab. Its growth will be 1>2>4>8>16>...This is called Geometric series. In the geometric series (progression) we multiply each term by some fixed number to get the next term. For example,
1,
3
(2.3) This expression is an example of an infinite series,
2.3
and we are asked to find its sum. Not all infinite series have sums.
Let us find the sum of n terms in ( ). The formula for the sum of n terms of the geometric progression is
(2
Using .3) ,
a + ar + ar
2+ ar
3+ …+ ar
n-1+ …
Series such as (2.3) whose terms form a geometric progression are called geometric series. We can write a geometric series in the form
S
the sum of the geometric series is by definition S lim
nS
n4 16
9
2 8
c) a, ar, ar
2, ar
3, ….
are geometric progressions. Let us consider the following expression
The geometric series has a sum if and only if r 1 , and in this case the sum is S
Repeating decimals
A repeating decimal can be thought of as a geometric series whose common ratio is a power of
1101000 3 100
3 10
0.3333... 3
3 1 9 10 10
3 10
1 1 10
3 1 r
0.3333... a
Try now : 0.6666... , 0.56565....
.
...
If we have the series a
na
1 a
2 a
3 a
4 ... a
n ...
Now consider the sums S
nthat we obtain by adding more and more terms of the series.
We define
S
1 a
1S
2 a
1 a
2S
3 a
1 a
2 a
3S
n a
1 a
2 a
3 . . . a
nEach S
nis called a partial sum; it is the sum of the first n terms of the series. as n increases, the partial sums may increase without any limit as in the series (2.1a).
1
2 2
2 3
2 4
2 ...
- 3 -
If a series has a finite sum, it is called convergent. Otherwise it is called divergent.
We have been talking about series which have a finite sum. We have also seen that there are series which do not have finite sums, for example (2.1 a).
It is understood that S is a finite number. If this happens, we make the following definitions:
(1) If the partial sums S
nof an infinite series tend to a limit S, the series is called convergent. Otherwise it is called divergent.
(2) The limiting value S is called the sum of the series.
lim
nR
n lim
n S S
n S S 0
(3) The difference R
n= S-S
nis called the remainder. From equation (2.4), we see that
3. Convergent and Divergent Series
Let S=1+2+4+8+16+...
2S=2+4+8+16+32+... = S-1 2S=S-1 > 2S-S=-1 > S=-1 ? ? ?
lim
nS
n S (2.4)
4. Testing Series for Convergence; The Preliminary Test
Preliminary test.This is not a test for convergence; the preliminary test can never tell you that a series converges. For example, the harmonic series
The terms are tending to 1, so by the preliminary test, this series diverges and no further testing is needed.
1 1
2 1
3
1 1
4
1 . . . 5
If lim
na
n 0 , we must test further.
If lim
na
n 0
, the series is divergent .2 1
3
2 4
4
3 . . .
5
n 1n 1
n
1
n
1 n
= 0 On the other hand, in the series
The nth term certainly tends to zero, but we shall soon show that the series
5. Tests for Convergence of Series of Positive Terms; Absolute Convergence A. The Comparison test:
The terms of the sequence an are compared to those of another sequence bn. if, For all n,
For all n,
for convergence.
2 1
2 1
4
1 1
8 1
n
...
n 1
16
n!
1 1 1
2
1 1
6
1 ...
n 1
24
n1
a
n diverges.
b
ndiverges, then so
n1
a
nb
n, and
n1
a
nconverges.
b
nconverges, then so
n1
0 a
nb
n, and
As a comparison series, choose the geometric series Example. Test
Solution
are smaller than the corresponding terms of
n 1
1 n!
1
2 . Therefore
for all
n3. We know that the geometric series converges because its ratio is 1
n1
2
n
n 1n!
1
In our example, the terms of
Notice that we do not care about the first few terms in a series, because they can affect the sum of the series but not whether it converges.
converges also.
- 5 -
We can use this test when the terms of the series are positive and not increasing, that
not hold for a finite number of terms.
The integral test states that:
is infinite → a
nis divergent series
If
andn
is finite → a
nis convergent series
If
andnB. The Integral test:
Using the integral test, we evaluate
Since the integral is infinite, the series diverges.
1 n dn ln n
1
2 1
3
1 1
4
1 . . . 5
C. Ratio Test
We consider another test which will handle many cases in which we cannot evaluate the integral.
inconclusive, and the series may converge or diverge.
n 1 n
lim a r
a
Assume that for all n,
a
n 0
. Suppose that there exists r such thatThe integral test depends on your being able to integrate
a
ndn
; this is not always easy.1
2!
1 3!
1 4!
1 1
. . . . n!
n 1 0 lim 1
n 1 1 n 1 n n 1 . . . 3. 2.1
n n 1 . . . 3.2.1 n 1 !
n ! n!
1 n 1 !
1
n
So, the series converges.
n
If
r1, then the series converges. If r1, then the series diverges. If r1, the ratio isExample: Test for convergence the series Solution
Example. Test for convergence the harmonic series Solution
. . . n . . . 1 4 1 3 1 2 1
n n
n 1
1
n 1
n 1 n
lim n lim
n 1 1
1 n
1 1
Solution
Example: Test for convergence the series
1
Here the test tells us nothing and we must use some different test.
is, when a
n1 a
n. This test can still be used even if the condition a
n1 a
ndoes
- 6 - D. A Special Comparison Test
This test has two parts: (a) a convergence test, and (b) a divergence test.
(a) If
n1
b
nis a convergent series of positive terms and a
n 0 and a
nb
ntends to a finite limit, then
n1
a
n converges.(b) If
n1
d
nis a divergent series of positive terms and a
n 0 and a
nd
ntends to a limit greater than 0 (or tends to ), then
n1
a
ndiverges.
Example 1. Test for convergence
4 n 7 n 2 2 n
25n 1
n 3
3 2
n n
n
2
1
n 3 3
n 3
2
For large n, we find
2 n
2 5 n 1
is nearly2 n
2 to quite high accuracy. Similarly, the denominator is nearly4 n
3.Solution
n 1 4 n 7 n 2
2n 5 n 1 b lim
lim a
3 2 22 n
n n
n
4 n 7 n 2 n 2n 5 n 1
lim
3 22 2
n
n 2 n 4 7
n 1 n 2 5 lim
2 2 n
2
4
This is a finite limit, so the given series converges.
So we consider as a comparison series just
by comparing their logarithms since ln N and N increase or decrease together. We have
ln 3
n n ln 3
, andln n
3 3ln n
. Now ln n is much smaller than n, so for large n we have . It is clear that this series is divergent.Now by test (b)
Which is greater than zero, so the given series diverges.
n 2
5 2
n 3n
n 2 5
n ln 33 ln n, and
3
n n
3. Thus the comparison series is
n
n 3
5 2
n
5 n
3 n
3
lim n 5 n
3 n
n
3 1
lim 3
n
1 n 5 1
Solution3
nn
3n 5 n
Here we must decide which is the important term as n ; is it 3n or n3? We find out Example . Test for convergence
- 7 -
n 3n
3 n
3
3 n
5 1 6. Alternating Series
with an 0 (or an 0 ) for all n. Its terms alternate between positive and negative.
Like any series, an alternating series converges if and only if the associated sequence of partial sums converges.
1
na
(6.1)n 0
n
An alternating series is an infinite series of the form
7. Useful facts about series
We state the following facts:
1. The convergence or divergence of a series is not affected by multiplying every term of the series by the same constant. Neither is it affected by changing a finite number of terms (for example, omitting the first few terms).
(an + bn). The resulting series is convergent, and its sum is obtained by adding (subtracting) the sums of the two given series.
3. The terms of an absolutely convergent series may be rearranged in any order without affecting either the convergence or the sum.
bn may be added (or subtracted) term by term
n1an and
n12. Two convergent series
and find the limit
Hence the condition for the convergence of a power series is obtained as
where R is called the radius of convergence. At the end points the ratio test fails; hence these points must be analyzed separately.
n 1 n
n 1 n 1
n n
n 1
u
nu
a x
a x a
a x
Where the coefficients
a
n are independent of x. To use the ratio test we writef x a
0a
1x a
2x
2. . . a x
n nn 0
n
8. Power Series; Interval of Convergence
Series with their general term given as
u
n x a
nx
are called power series;x R R x R
n 1 n
n
lim a
a
R
1
We use the ratio test
lim
nn 1
1 n
1 n
u
nu
n 1x
n1 n
x n
n x
1 n
x
2 3 4 n1 2
x
3 x
4
x x
. .
. . . .
n
n 1
lim
R
n
n
n
1 1
lim 1
n
So the radius of convergence R is 1; thus the series converges in the interval
1 x 1
. On the other hand, at the end point x=1 it is divergent, while at the other end point, x=-1, it is convergent. So the interval of convergence is1 x 1
Example 1: Test convergence for the power series
The ratio gives
Thus the radius of convergence is zero. Note that this series converges only for x=0.
1 x 2!x
2 3!x
3 . . . n!x
n. . .
n1
n
a n 1 !
a n 1
n!
lim
n n 1
R 1
Example 2: Test convergence for the power series
- 8 -
we findand
Hence the radius of convergence is infinity. This series converges for all x values.
lim
n1 1
n 1 0
R
n 1
a
na
n 1 !
n! n 1 1
2
3
n 1 x 2!
x
3!
x x
. . . . . . n!
Example 3: Test convergence for the power series
1 2n 1 ! x . . .
. . . 7!
x 5!
x 3!
x x
n 1 2n 1
3 5 7
The ratio
2n 1 ! 2n 1 x 2n
x 2n 1 !
x u
u
2n 1 2n 1 2n n 1
2n 1 2n x 0
lim
2
n
Example 4: Test convergence for power series
So the radius of convergence is infinity.
This series converges for all x values.
The ratio
and
If x = -3, the series is
2 n
1 2
x 2
3
x 2 x 2
. . . . . .
n 1
n 1
n
u x 2
n 1 nu
x 2
n 2 x 2
n 1 n 2
n 1
lim
n x 2 n 1 x 2
n 2
The series converges for
x 2 1
; that is, for 1 x 2 1,
or 3 x 1
.. . . 4
1 3 1 2
1 1
n 0
n 1
. . . 1 3 1 2 1 1
This is divergent by the integral test. Thus the series converges for
3 x 1
Example 5: Test convergence for the power series
- 9 -
9. Theorems about Power Series
A power series may be differentiated or integrated term by term; the resulting series converges to the derivative or integral of the function represented by the original series within the same interval of convergence as the original series (that is, not necessarily at the endpoints of the interval).
Two power series may be added, subtracted, or multiplied; the resultant series converges at least in the common interval of convergence. You may divide two series if the denominator series is not zero at x=0. The resulting series will have some interval of convergence.
The power series of a function is unique, that is, there is just one power series of the form
n0an xn which converges to a given function.
One series may be substituted in another provided that the values of the substituted series are in the interval of convergence of the other series.
This is convergent by the alternating series test. For x = -1, the series is
1. Addition and subtraction
When two functions f and g are decomposes into power series around the same center c, the power series of the sum or difference of the functions can be obtained by termwise addition and subtraction. That is, if
Then
g x b
nx c
n n 0
f x a
nx c
n n 0
f x g x a
nb
nx c
n n 0
