Class notes Oct. 2nd
Some examples on series
We begin with some comparison criteria useful in deciding the convergence of series.
Theorem 1. (Theorem 2.7.4. in your text) Let (ak ),(bk ) be sequences satisfying
0  ak  bk for all k  N . Then:

(i)
If
 bk converges, then
k 1

(ii)
If
a
k 1

a
k 1
k
converges as well.

k
diverges, then
b
k 1
k
diverges.
Theorem 2. Let (ak ),(bk ) be sequences satisfying ak  0, bk  0 for all k  N , and


a
suppose that lim n  L, L  0 . Then  ak and  bk either both converge or both
n  b
k 1
k 1
n
diverge.
Remark 1. This is a very useful criterion in deciding whether a series converges or
diverges since the analysis of a series that seems more complicated can be reduced to a
simpler one which is equivalent to it from the point of view of convergence. Notice that
nothing is said about the value of the infinite sum. Below you will see how this result can
be applied.
a
Proof: Since lim n  L, L  0 we have that for any   0 there exists
n  b
n
a
n0  N :| n  L |  , for all n  n0 . We will not need to work with any   0 , so we will
bn
a
choose a particular value,   1 . Then we have | n  L | 1, for all n  n0 , or equivalently
bn
bn  an  bn L  bn , for all n  n0 , or ( L  1)bn  an  bn (1  L), for all n  n0 (1) .
But since L  0 , from the algebraic properties of limits of sequences we have that the
b
1
following limit exists: lim n  . Then, as above, we obtain that there exists a rank
n  a
L
n
n1  N such that ( L  1)an  bn  an (1  L), for all n  n1 (2) .

Let us now assume that the series
a
k 1
k
converges. The idea of the proof is the
following: we use inequality (2) which gives us that from a certain rank on all the terms

in the series
b
k 1
k
are below a constant multiple of the corresponding terms in the series

a
k 1
k
. Since the convergence of a series is not affected by what happens to a finite

number of terms the series
b
k 1
k
will also be convergent. Now, for a rigorous proof, we

will show that the sequence of partial sums of the series
b
k
k 1
Since
is Cauchy.

m
k 1
k 1
 ak converges, the sequence of partial sums, Sm   ak is Cauchy, so for any

  0 there exists l  0 such that | Sm  S n |
, for all m, n  l . Without loss of
1 L
generality we may consider l  n1 (that is, if l  n1 great, if l  n1 take n1 instead of l, the
Cauchy property will still hold).
m
Let m, n  l , and suppose n  m . Denote Tm   bk . Then
k 1
n
m
k 1
k 1
| Tm  Tn |  bk   bk 
n

k  m 1
bk 
n

k  m 1
ak (1  L) 

1 L
(1  L)   ,

and therefore the sequence of partial sums of the series
b
k 1
k
is Cauchy, so the series
converges.

Let us now assume that the series
a
k 1
k
diverges. That means the partial sums grow
without bound, so for any M  0 there exists l  0 , and we may also choose it l  n0 ,
m
such that Sm   ak  M , for all m  l . We want to use inequality (1) which guarantees
k 1
that the terms bk stay above some constant multiple of the terms ak , for k  n0 . The idea
is again that we can ignore what happens to the terms for k  n0 , but in order to carefully
m
n0
n0
k 1
k 1
write this we have for the given M above, Sm   ak  (1  L)M   ak  (1  L) bk ,
k 1
for all m  l , which is equivalent to
n0
m

k  n0 1
ak  (1  L) M  (1  L) bk , for all m  l .
k 1
m
We now use inequality (1). We have

k  n0 1
bk 
n0
1
a

M

bk , for all m  l .
 k

1  L k  n0 1
k 1
m
m
But this last inequality is equivalent to Tm   bk  M , proving that the series
k 1

b
k 1
k
diverges.
QED
Remark 2. The theorem does not give any conclusion when the limit of the ratio
a
lim n  0 . We will se in some of the examples that follow that in that case there is no
n  b
n
definite conclusion.

Example 1. Prove that the series
1
 k ! converges.
k 1
1
1
1
1
. We have ak 

 for all k  1 . Therefore
k!
k (k  1) k  1 k
n
n
1
1
1 1 1
1
1
1
Sn   ak  1   (
 )  1  1     ... 
  2 .
k
2 2 3
n 1 n
n
k 1
k 2 k  1

1
1
Therefore, the partial sums of the series 
are bounded above by 2  , and since the
n
k 1 k !
series has positive terms, by Theorem 1 it follows that it is convergent.

a
1
1
k2
Notice that if we denote bk  2 , lim k  lim  0 , and as we know the series  2
k k  bk k  k !
k 1 k
converges.

a
1
k
1
However, if we define ck  , lim k  lim  0 , but the series  diverges.
k k  ck k  k !
k 1 k
a
This is why Theorem 2 cannot give any conclusions when lim n  0 .
n  b
n
Let ak 

Example 2. Determine whether the series
(
k  1  k ) converges or diverges.
k 1
(k  1)  k
1
1
. At this


k 1  k
k 1  k 2 k 1

1
moment we can see that the series diverges because the series 
diverges. Therefore
k
k 1


1
1
we can just compare 
with 
using Theorem 2.
k
k 1 2 k  1
k 1
Let ak  k  1  k . We have ak 
1
1
k
1

 lim
 , so the series
n  2 k  1
n

k
2 k 1 2
We have lim

2
k 1
1
diverges, and
k 1

then the series
a
k 1
k
diverges as well.
Before looking at more examples, here is an important result which helps decide in
certain cases when a series diverges.

Theorem 3. (Theorem 2.7.3.) If a series
a
k 1
k
converges, then lim an  0 .
n 
Proof: Since the series converges, the sequence of partial sums is a Cauchy sequence. We
will write this fact with a particular choice as follows: for each   0 there exists n0  0
such that | Sm  Sn |  , for all m, n  n0 . In particular, if m  n  1, | Sn1  Sn || an1 |  ,
and therefore lim an  0 .
n 
QED
Remark 3. Use this result as a divergence criterion rather than as a convergence

criterion. If for a series
a
k 1
k
, lim an  0 , this will not imply that the series converges.
n 
On the other hand, if lim an  0 or it does not exists, then the series diverges.
n 

Example 3. Determine whether the series
k
 k  1 converges or diverges.
k 1
n
 1 , therefore the series diverges.
n 1

k
Example 4. Determine whether the series  2
converges or diverges.
k 1 k  2k  2
We have lim
n 
k
. A quick evaluation of the term ak shows that the numerator is of
k  2k  2
order k 1/ 2 , whereas the dominant term in the denominator is of order k 2 , so ak is of
1
order k 3/ 2 . We then try to use Theorem 2 with bk  3/ 2 . We have
k
3/ 2
a
k k
k2
lim k  lim 2
 lim 2
1.
k  b
k  k  2k  2
k  k  2k  2
k
Let ak 
2

Then, since
k
k 1
1
3/ 2

converges, Theorem 3 gives that
k
k 1
2
k
converges.
 2k  2