MA 124C – Discussion #10 – Sections 8.5
Mixed Practice
a. Converges - Ratio Test
b. Diverges - Divergence Test
c. Converges - Geometric Series
d. Converges - p-series
e. Converges - Integral Test + p-series
f. Converges - Ratio Test
g. Diverges - Limit Comparison Test
h. Diverges - Divergence Test
i. Diverges - Divergence Test
j. Diverges - Integral Test + p-series
k. Converges - Ratio Test
l. Converges - Limit Comparison Test
8.6
1.
Alternating Series
Sk 1  (1)k 1 ak  Sk and the terms of the series, (1)k 1 ak , alternate signs, so if (1)k 1 ak
k 1
k 1
is positive, then Sk  Sk 1  (1) ak  Sk 1 (increases), but if (1) ak is negative, then
2.
Sk  Sk 1  (1)k 1 ak  Sk 1 (decreases), therefore the sequence of partial sums is neither
strictly increasing, nor strictly decreasing.
a. 0  ak 1  ak for all k greater than some index N (terms are non-increasing in
3.
magnitude)
lim ak  0
b. k
a. Converges
4.
5.
6.
b. Converges
c. Diverges
d. Converges
Rn  an1 . This makes sense because the limit of the sequence of partial sums, S, must
fall between any two consecutive terms, Sn and Sn1 , so Rn  S  Sn  Sn1  Sn  an1 .
n1
n1
If (1) an1 is negative, then Sn is an overestimate. If (1) an1 is positive, then Sn is
an underestimate.
a. S4
b. S6
a.

ak 
 (1)k
k1
2
1
k2
c. S499
is a decreasing sequence and
 1  14  19 
lim k12  0
k
b. S3
c.
b. S2
c.
31
36
k 1
7.
a.

ak 
1
(2k 1)!
is a decreasing sequence and
1
1
19
 (2(1)
k 1)!   3!  5!   120
k 1
k
lim (2k11)!  0
k
8.6
Absolute Convergence
1.
a.
a
a
 a converges.
a
a
converges conditionally, if  converges but 
k
b.
converges absolutely, if
k
k

2.
k
does not converge.
k1
 (1)k
k 1
a.
k

is the Alternating Harmonic Series, it converges conditionally.
k1
 (1)k
2
b.
3.
4.
k 1
converges absolutely.
a
a
a
a
If  k converges, then  k converges. If  k diverges, then  k diverges.
a. Converges Absolutely (Root Test)
b. Converges (AST), but not absolutely convergent (p-Series)  Conditionally
Convergent
c. Absolutely Convergent (p-Series)
d. Diverges by Divergence Test
1
k3
5.
e. Absolutely Convergent (Comparison to )
1  13  312  313  314  315 
a. The terms of
… are not all nonnegative and they alternate in
pairs.

b. Since

6.
7.
 31
k 1
k
1
is a Geometric Series with r  3  1 , the series converges absolutely.
k
 k(1)
ln k
converges (AST), not absolutely convergent (Integral Test)  Conditionally
Convergent
k2
a. For p  1 , converges absolutely by p-Series
b. For 0  p  1 , converges by AST, but not absolutely by p-Series  Conditionally
Convergent c. For p  0 , diverges by Divergence Test.
Mixed Practice
1.
Divergence Test
2.
The interior terms cancel and we can express the nth partial sum in simplified form, then
find its limit. For a telescoping series, the ak is often a difference of two fractions or is a
fraction that can be broken up through the method of partial fractions.
3.
If there is a continuous, positive, decreasing function such that f (k)  ak and we can

easily integrate f, then

4.
a.
k
b.
5.
 ar
k0
kc
converges if and only if


c
f (x) dx
converges.
p
is a p-Series, it converges for p  1 and diverges for p  1 .
k 1

 ak
k
is a geometric series, it converges to
a
1r
for r  1 and diverges for r  1 .
k
k
a. Ratio Test if terms contain k! , k , a . b. Root Test if terms contain k in exponent.
c. Comparison for rational function when it’s easy to establish the necessary inequality
d. Limit Comparison for a rational function without establishing an inequality
k sin k
 2 , or cos(k )
e. Alternating Series when the terms contain (1) ,
6.
Diverges by Divergence Test
7.
Converges to –1, Telescoping Series
8.
Diverges by Integral Test
9.
Converges by Limit Comparison to p-Series p  2
lim x x1   lim 1 
x
x
x

1 x
x 1
 lim 1  1x   e1  1
x
x
10.
Converges by Ratio Test (Hint:
11.
Diverges by Comparison or Limit Comparison with Harmonic series
12.
Diverges by Limit Comparison with p-Series p  2
13.
Converges to
14.
Converges by Root Test
15.
Diverges by Root Test
1
10
, Geometric Series
)