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The sum of the infinite and finite geometric sequence
2
The sum of the first n terms of a sequence is represented by summation notation.
index of summation
upper limit of summation
lower limit of summation
The sum of a finite geometric sequence is given by
5 + 10 + 20 + 40 + 80 + 160 + 320 + 640 = ?
n = 8
a1 = 5
The sum of the terms of an infinite geometric sequence is called a geometric series.
a1 + a1r + a1r2 + a1r3 + . . . + a1rn-1 + . . .
If |r| < 1, then the infinite geometric series
has the sum
Example: Find the sum of
The sum of the series is
Convergent and Divergent Series
Convergent and Divergent Series
If the infinite series has a sum, or limit, the series is convergent.
If the series is not convergent, it is divergent.
Ways To Determine Convergence/Divergence
1. Arithmetic – since no sum exists, it diverges
2. Geometric: If |r| > 1, diverges If |r| < 1, converges since the sum
exists3. Ratio Test (discussed in a few
minutes)
Example
Determine whether each arithmetic or geometric series is convergent or divergent.
1/8 + 3/20 + 9/50 + 27/125 + . . . r=6/5 |r|>1 divergent
18.75+17.50+16.25+15.00+ . . . Arithmetic series divergent
65 + 13 + 2 3/5 + 13/25 . . . r=1/5 |r|<1 convergent
Other Series
When a series is neither arithmetic or geometric, it is more difficult to determine whether the series is convergent or divergent.
Ratio Test
In the ratio test, we will use a ratio of an and an+1 to determine the convergence or divergence of a series.
1
nIf lim 1, the absoconverg lutelyesn
n
n
aa
a
1
nIf lim 1, th diverges.e n
n
n
aa
a
1
nIf lim 1, the ratio test i incos nclusive.n
n
a
a
Review:
2
2nlim
n ba
nd
n c
e
2
3n0lim
an bn c
dn e
Leading coefficient is
a
Leading coefficient is
d
Denominator degree is greater
a
d
Test for convergence or divergence of:
2
3
n
na n
1
2
3
n
n
n
1
12
( 1)3n
n
na
1
1
2( 1)
2
3
3
nn
n
n
na
an
11 2
3
n nn
n
1 2
3
n
n
1lim n
nn
a
a
2 1lim
3 n
n
n
2
3 Since this ratio is less
than 1, the series converges.
Test for convergence or divergence of:
2
2n n
na
2
11
( 1)
2nn
na
2
2
11
( 1)2
2
nn
nn
na
na
2
1 2
1 2
2
n
n
n
n
2
2 1
( 1) 2
2
n
n
n
n
1lim n
nn
a
a
Since this ratio is less than 1, the series converges.
2
1 2nn
n
2
2
1 2 1
2
n n
n
2
2
1 2 1lim
2 n
n
n
n
The ratio of the leading coefficients
is 1
1
2
Test for convergence or divergence of:
( 2)
( 1)n
na
n n
1
( 3)
( 1)( 2)n
n
na
n
1
( 3)( 1)( 2)
( 2)( 1)
n
n
nn nanan n
( 3) ( 1)
( 1)( 2) ( 2)
n n n
n n n
1lim n
nn
a
a
Since this ratio is 1, the test is inconclusive.
Coefficient of n2 is 1
1
1
( 1) ( 2)
( 1)
n
n
n
n n
2
( 3)
( 2)
n n
n
2
( 3)lim
( 2)n
n n
n
Coefficient of n2 is 1
1
Example
Use the ratio test to determine if the series is convergent or divergent.
1/2 + 2/4 + 3/8 + 4/16 + . . .
1 1
1
2 2_ _n nn n
n na and a
Since r<1, the series is convergent.
1
1
11 2 12lim lim lim 1/ 2
2 22
nn
nn n nn
nn n
rn n n
Example
Use the ratio test to determine if the series is convergent or divergent.
1/2 + 2/3 + 3/4 + 4/5 + . . .
1_1 1
1 ( 1) 2_
1n n
n n na and a
n n n
Since r=1, the ratio test provides no information.
2
2
1 1 1 2 12lim lim lim 12 2
1n n n
n n n n nnrn n n n nn
Example
Use the ratio test to determine if the series is convergent or divergent.
2 + 3/2 + 4/3 + 5/4 + . . .
1
1 ( 1) 1
1_ _
2
1n n
n n na and a
n n n
Since r=1, the ratio test provides no information.
2
2
2 2 21lim lim lim 11 1 1 2 1n n n
n n n n nnrn n n n n
n
Example
Use the ratio test to determine if the series is convergent or divergent.
3/4 + 4/16 + 5/64 + 6/256 + . . .
12 2( 1) 2 2_ _
2 ( 2) 1 3
2 2 2n nn n n
n n na and a
Since r<1, the series is convergent.
22 2
2 2 22
33 2 3 12lim lim lim
2 2 2 2 ( 2) 42
nn
nn n nn
nn n
rn n n
Example
Use the ratio test to determine if the series is convergent or divergent.
1_ _1 1
! ( 1)!n na and an n
Since r<1, the series is convergent.
11 ! 1( 1)!
lim lim lim 01 ( 1)! 1 1
!n n n
nnr
n nn
1 1 11 ....
1 2 1 2 3 1 2 3 4
Example
Use the ratio test to determine if the series is convergent or divergent.
1
1
2 2
( 1) ( 1)(_
2_
)
n n
n na and an n n n
Since r>1, the series is divergent.
112
2 ( 1) 2( 1)( 2)lim lim lim 2
2 ( 1)( 2) 2 2( 1)
nn
n nn n n
n n nn nr
n n nn n
2 4 8 16....
1 2 2 3 3 4 4 5
