=0
(

1)
n
√
n
, which converges by Leibnitz’ theorem. For
x
= 2, we get
∑
∞
n
=0
1
√
n
which diverges (see this either by the integral test or comparison with the harmonic series).
Combining this together, we get that the interval of convergence is [0
,
2), and the radius of
convergence is 1.
Example 51.
Find the interval of convergence and radius of convergence of
∑
∞
n
=1
n
√
n
(2
x
+ 5)
n
Solution.
First, we have to express the given function as a power series:
∞
X
n
=1
n
√
n
(2
x
+ 5)
n
=
∞
X
n
=1
n
√
n
(2
n
)
x
+
5
2
n
(in particular, note that
a
=

5
2
, not

5). Using thenth root test, we see
(1)
n
and if
x
+
5
2
=

1
2
,
then
x
=

3 and the corresponding series is
∑
∞
n
=1
n
√
n
(

1)
n
. Neither of these series has
a
n
→
0, so
they both diverge. This means that the interval of convergence is (

3
,

2).
7.2.
TermByTerm Calculus.
If
∑
c
n
(
x

a
)
n
converges on some positive radius of convergence
(i.e. it doesn’t diverge for every
x
6
=
a
), then
f
(
x
) =
∑
c
n
(
x

a
)
n
is a continuous function (not to
difficult to show), and we can reduce taking derivatives and integrals of
f
to taking derivatives and
integrals of the terms (more difficult to show):
7.2.1.
TermByTerm Differentiation.
Theorem 7.3.
If
f
(
x
) =
∑
∞
n
=0
c
n
(
x

a
)
n
converges on

x

a

< R
, then:
(1)
f
has derivatives of arbitrary order on
(
a

R, a
+
R
)
(i.e.
f
0
(
x
)
, f
00
(
x
)
, f
000
(
x
)
, . . .
exist)
(2)
The derivatives are given by:
f
0
(
x
) =
∞
X
n
=1
nc
n
(
x

a
)
n

1
, f
00
(
x
) =
∞
X
n
=2
(
n
)(
n

1)(
x

a
)
n

2
, . . .
on
(
a

R, a
+
R
)
.
We can use this to express some power series in other ways:
Example 52.
Recall that
f
(
x
) =
∑
∞
n
=0
x
n
converges for

1
< x <
1
. The above theorem tells us
that
f
0
(
x
) =
∞
X
n
=1
nx
n

1
and
f
00
(
x
) =
∞
X
n
=2
n
(
n

1)
x
n

2
On the other hand, we know that
f
(
x
) =
1
1

x
, so
f
0
(
x
) =
1
(1

x
)
2
and
f
00
(
x
) =
2
(1

x
)
3
. This gives us
formulae:
∞
X
n
=1
nx
n

1
=
1
(1

x
)
2
,
and
∞
X
n
=2
n
(
n

1)
x
n

2
=
2
(1

x
)
3
It is important to note here that
c
n
don’t involve
x
in any way (otherwise the series would not
be a power series). It is not true that if we take any series
∑
f
n
(
x
) of functions of
x
that we can
differentiate termbyterm as above:
Example 53.
The series
∑
∞
n
=1
sin(
n
!
x
)
n
2
converges absolutely for all
x
, but if we differentiate term
byterm we get
∑
∞
n
=1
n
! cos(
n
!
x
)
n
2
which diverges for all
x
.
MATH 214: SEQUENCES AND SERIES
13
7.2.2.
TermByTerm Integration.
Theorem 7.4.
Given
f
(
x
) =
∑
∞
n
=0
c
n
(
x

a
)
n
that converges on

x

a

< R
, the function
F
(
x
) =
∑
∞
n
=0
c
n
(
x

a
)
n
+1
n
+1
converges on

x

a

< R
and
Z
f
(
x
) d
x
=
F
(
x
) +
C
We can use this to identify series with other functions as well:
Example 54.
Identify
f
(
x
) =
∑
∞
n
=0

(1

t
)
n
+1
n
+1
on
0
< x <
2
.
Solution.
First, differentiate the given series termbyterm to get
f
0
(
x
) =
∞
X
n
=0
(1

t
)
n
=
1
1

(1

t
)
=
1
t
This means that
f
(
x
) =
R
f
0
(
x
)d
x
+
C
= ln(
x
)+
C
. Setting
x
= 1 we get
C
= 0, i.e.
f
(
x
) = ln(
x
).