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17. Power series 191
Cauchy–Hadamard formula for the radius
of convergence of a power series
Formulation of the Cauchy–Hadamard
theorem 3.15A/08:30 (04:04)
Theorem (Cauchy–Hadamard theorem on the radius of con-
vergence of a power series).
The radius R of convergence of the power series
P
∞
n=0
cnxncan be found
by the formula
R =
1
α
, α = lim
n→∞
n
p
|cn|. (5)
It is assumed that R = 0 if α = +∞ and R = +∞ if α = 0. Formula (5)
is called the Cauchy–Hadamard formula.
Remark.
In formula (5), the coefficients cnare considered starting from n = 1, since
a root of degree zero is not defined.
Proof of the Cauchy–Hadamard theorem 3.15A/12:34 (13:25)
We give a proof for the special case when the sequence {|cn|} has a usual
limit (finite or infinite). The proof for the general case is given, for example,
in [4, Ch. 11, Sec. 380].
So, suppose the sequence {|cn|} has a limit α:
lim
n→∞
n
p
|cn| = α ∈ R ∪ {+∞}. (6)
Let us consider three cases.
1. 0 < α < +∞. In this case, formula (5) does not require special
interpretations.
We want to prove two facts: if |x| < R, then the initial series
P
∞
n=0
cnx
n
converges, if |x| > R, then the initial series diverges.
We choose some point x0and consider the numerical series
P
∞
n=1
|cnx
n
0
|
denoting its common term by an: an= |cnx
n
0
|.
To study the convergence of this series, we use the limit Cauchy test.
Recall its formulation: if there exists a limit lim
n→∞
n
√
an= q for the series
P
∞
n=1
an, then the series converges if q < 1 and diverges if q > 1.
Let us find the value of q, provided that an= |cnx
n
0
| and the limit rela-
tion (6) holds:
q = lim
n→∞
n
q
|cnx
n
0
| = lim
n→∞
|x0|
n
p
|cn| = |x0| lim
n→∞
n
p
|cn| = |x0|α.
192 M. E. Abramyan. Lectures on integral calculus and series theory
Thus, q = |x0|α. Therefore, if |x0| < R, i. e., |x0| <
1
α
, then q = |x0|α < 1
and, by the limit Cauchy test, the series
P
∞
n=1
|cnx
n
0
| converges. If |x0| > R,
then q = |x0|α > 1 and, by the same test, the series
P
∞
n=1
|cnx
n
0
| diverges.
Thus, the value R =
1
α
is the radius of convergence of the series
P
∞
n=1
|cnx
n
0
|, and the same result is true for the initial series
P
∞
n=0
cnx
n
0
, since,
according to the second Abel theorem, the power series converges absolutely
on the convergence interval.
2. α = 0. We show that in this case R = +∞, i. e., the initial series
converges at any point x ∈ R.
We choose some point x0∈ R, consider the numerical series
P
∞
n=1
|cnx
n
0
|,
and find the value q for it:
q = lim
n→∞
n
q
|cnx
n
0
| = lim
n→∞
|x0|
n
p
|cn| = |x0| lim
n→∞
n
p
|cn| = 0.
Thus, q = 0 < 1 and, by the limit Cauchy test, the series
P
∞
n=1
|cnx
n
0
|
converges for any point x0∈ R. So, the initial series
P
∞
n=0
cnx
n
0
converges
absolutely for any point x0∈ R and therefore R = +∞.
3. α = +∞. We show that in this case R = 0, i. e., the initial series
diverges at any point x 6= 0.
We choose some point x06= 0, consider the numerical series
P
∞
n=1
|cnx
n
0
|,
and find the value q for it:
q = lim
n→∞
n
q
|cnx
n
0
| = lim
n→∞
|x0|
n
p
|cn| = |x0| lim
n→∞
n
p
|cn| = +∞.
Taking into account the remark on the limit Cauchy test in the case
q = +∞, we obtain that the series
P
∞
n=1
|cnx
n
0
| diverges for any point x06= 0.
So, the initial series
P
∞
n=0
cnx
n
0
also diverges for any point x06= 0 and there-
fore R = 0.
Examples of application
of the Cauchy–Hadamard formula 3.15A/25:59 (09:37)
1. Consider the power series
P
∞
n=1
x
n
n
. In this case, cn=
1
n
. Since
lim
n→n
n
q
1
n
= 1 (see the convergence theorem for the sequence {
n
√
n} in
[1, Ch.5]), we obtain that α = 1, R = 1. Therefore, this series converges
absolutely for |x| < 1 and diverges for |x| > 1.
Let us analyze the convergence of this series for |x| = 1. In the case of
x = 1, we get the harmonic series
P
∞
n=1
1
n
, which is divergent. In the case of
x = −1, we get a convergent alternating series
P
∞
n=1
(−1)
n
n
.
Thus, the domain of convergence of the power series
P
∞
n=1
x
n
n
is the half-
interval [−1, 1).
17. Power series 193
2. Consider the power series
P
∞
n=1
x
2n
2n
. In this case, the sequence {cn} is
not convergent. Indeed, let us write out the initial terms of this series:
∞
X
n=1
x
2n
2n
=
x
2
2
+
x
4
4
+
x
6
6
+ .. .
We obtain that the coefficient c1(i. e., the coefficient of the first power
of x) is 0, the coefficient c2is
1
2
, the coefficient c3is 0, and so on. The formula
for the coefficients cntakes the form
cn=
1
n
, n = 2k,
0, n = 2k − 1, k = 1, 2, . . .
Therefore, the sequence
n
√
c
n
contains the following elements:
{
n
√
cn} =0,
1
√
2
, 0,
1
4
√
4
, 0,
1
6
√
6
, . . ..
This sequence has no limit, since there exist an infinite number of elements
of this sequence in any neighborhood of points 0 and 1. We can also say that
the sequence
n
√
c
n
has two partial limits: 0 and 1.
However, according to the general Cauchy–Hadamard formula, we can
determine the radius of convergence of a power series if we find the limit
superior of the sequence
n
√
c
n
, which always exists. In our case, the limit
superior is 1; it is the limit of a subsequence containing elements with even
indices:
lim
n→∞
n
√
cn= lim
k→∞
1
2k
√
2k
= 1.
Therefore, by the Cauchy–Hadamard theorem, the radius of convergence
of the series
P
∞
n=1
x
2n
2n
is equal to 1.
Note that this series diverges at both endpoints of the convergence interval
(−1, 1), since, for the value x = 1 and for the value x = −1, we get the same
series
P
∞
n=1
1
2n
, which differs from the harmonic series only by the factor
1
2
.
Remark.
The Cauchy–Hadamard formula can also be used to find the radius of
convergence R of power series of the form
P
n
k=0
ck(x − x0)k, where x06= 0,
since the radius of convergence is determined by the coefficients ckonly and
does not depend on the center x0.
194 M. E. Abramyan. Lectures on integral calculus and series theory
Properties of power series
Continuity of the sum of a power series 3.15A/35:36 (08:26)
In all the theorems of this section, we consider the power series
n
X
k=0
ckxk. (7)
We assume that series (7) has a radius of convergence R > 0 (the case
R = +∞ is also allowed). We denote the sum of this series by S(x). This
sum is defined for all x ∈ (−R, R).
Remark.
All the results obtained in this section remain valid for power series of the
form
P
n
k=0
ck(x−x0)k, where x06= 0. Recall that the convergence interval has
the form (x0−R, x0+ R) for such series, where the radius of convergence R
can be found by the Cauchy–Hadamard formula.
Theorem 1 (on the continuity of the sum of a power series).
The function S(x) is continuous on the convergence interval (−R, R).
Proof.
Let x0∈ (−R, R). Choose a segment [−R + ε, R − ε] containing the
point x0and nested in the convergence interval (−R, R):
x0∈ [−R + ε,R − ε] ⊂ (−R, R).
As ε, we can take
R−|x0|
2
. In the case R = +∞, we can consider an arbitrary
segment of the real axis containing the point x0.
Using the first and second Abel theorems, we obtain that series (7) converges uniformly on the segment [−R + ε, R − ε]. In addition, the general
term of the series (7) has the form ckxkand therefore is a continuous function. These two facts imply, by virtue of the theorem on the continuity of
the uniform limit, that the limit function S(x) is continuous on the segment
[−R + ε, R − ε]; therefore, it is continuous at the point x0belonging to this
segment.
Since the point x0∈ (−R, R) was chosen arbitrarily, we obtain that the
function S(x) is continuous on the entire convergence interval (−R, R).
Integration of a power series 3.15B/00:00 (13:22)
Theorem 2 (on the integration of a power series).
Series (7) can be term-by-term integrated on any segment nested in the
convergence interval (−R, R), i. e., to find the integral of the sum S(x) of
17. Power series 195
series (7), it suffices to find the sum of the series whose terms are the integrals
of the terms of the initial series. Moreover, the radius of convergence of the
integrated series coincides with the radius of convergence of the initial series.
Proof.
First, we prove that the radius of convergence R0of the integrated series
is equal to the radius of convergence R of the initial series (7).
Let |x| < R. For definiteness, we will perform integration from 0 to x.
Consider the following series obtained by term-by-term integration of series (7):
∞
X
k=0
Z
x
0
cktkdt =
∞
X
k=0
c
k
k + 1
x
k+1
= c0x +
c
1
2
x2+
c
2
3
x3+ .. .
Denote by dkthe coefficient of xkin the resulting series: dk=
c
k−1
k
.
To find the radius of convergence R0of the integrated power series, we use
the Cauchy–Hadamard formula:
1
R
0
= lim
n→∞
n
p
|dn| = lim
n→∞
n
r
|c
n−1
|
n
=
lim
n→∞
n
p
|c
n−1
|
lim
n→∞
n
√
n
.
In the denominator, we indicated the usual limit, since the sequence {
n
√
n}
is convergent. The limit of this sequence is 1. We transform the numerator
as follows:
lim
n→∞
n
p
|c
n−1
| = lim
n→∞
|c
n−1
|
1
n−1
n−1
n
= lim
n→∞
n−1
p
|c
n−1
|
n−1
n
.
By the Cauchy–Hadamard formula, we get lim
n→∞
n−1
p
|c
n−1
| =
1
R
, the
limit of the exponent
n−1
n
, as n → ∞, is 1. Thus, the limit of the numerator
is
1
R
and finally we get
1
R
0
= lim
n→∞
n
p
|dn| =
1
R
.
So, we have proved that the radii of convergence R0and R coincide.
To complete the proof of the theorem, it remains for us to prove that the
sign of the integral can be moved under the sign of an infinite sum:
Z
x
0
S(t) dt =
Z
x
0
∞
X
k=0
ckt
k
dt =
∞
X
k=0
Z
x
0
cktkdt.
This formula is valid by virtue of the corollary on the integration of a uniformly converging functional series, since all conditions of this corollary are
fulfilled: the initial series (7) converges uniformly on the segment [0, x] nested
in the convergence interval (−R, R) and, in addition, all terms of the initial
series (7) are continuous functions on this segment.
196 M. E. Abramyan. Lectures on integral calculus and series theory
Differentiation of a power series 3.15B/13:22 (12:46)
Now we turn to the differentiation operation and consider the series ob-
tained by term-by-term differentiation of the initial series (7):
∞
X
k=0
(ckxk)0=
∞
X
k=1
kckx
k−1
. (8)
In this case, the summation starts with k = 1, since when differentiating the first term c0, which does not depend on x, we get 0. If dkis the
coefficient for the degree xkof the differentiated series, then it follows from
formula (8) that d
k−1
= kck, whence dk= (k + 1)c
k+1
. Let us find the radius
of convergence R0of the power series (8) by the Cauchy–Hadamard formula:
1
R
0
= lim
n→∞
n
p
|dn| = lim
n→∞
(n + 1)|c
n+1
|
1
n
=
= lim
n→∞
n+1
√
n + 1
n+1
n
· lim
n→∞
n+1
p
|c
n+1
|
n+1
n
. (9)
The first limit on the right-hand side of (9) is 1, since both the limit
of the base
n+1
√
n + 1 and the limit of the exponent
n+1
n
are 1 as n → ∞.
For the second limit, we obtain that the exponent
n+1
n
approaches 1 and
lim
n→∞
n+1
p
|c
n+1
| =
1
R
by the Cauchy–Hadamard formula, where R is the
radius of convergence of the initial series (7). Thus, the right-hand side of
equality (9) approaches
1
R
.
So, we have proved that
1
R
0
=
1
R
. This means that series (8) obtained by
formal differentiation of the terms of the initial series (7) has the same radius
of convergence as the initial series.
Theorem 3 (on the differentiation of a power series).
Series (7) can be term-by-term differentiated at any point in the convergence interval (−R, R), i. e., to find the derivative of the sum S(x) of the
series (7), it suffices to find the sum of the series whose members are the
derivatives of the terms of the initial series. Moreover, the radius of convergence of the differentiated series coincides with the radius of convergence of
the initial series.
Proof.
We have already proved the statement about the coincidence of the convergence radii.
It remains to prove that the sign of differentiation can be moved under the
sign of an infinite sum:
17. Power series 197
S0(x) =
∞
X
k=0
ckx
k
0
=
∞
X
k=0
ckx
k
0
=
∞
X
k=1
kckx
k−1
.
This statement is true by virtue of the corollary on the differentiation
of the functional series, since all conditions of this corollary are fulfilled: the
formally differentiated series (8) uniformly converges on any segment nested in
the convergence interval (−R, R), all the terms of the series (8) are continuous
functions, and the initial series (7) also converges uniformly on this segment
(note that, in the indicated corollary, it was only required that the initial
series converge at one point).
After differentiating the series (8) composed of differentiated members
of the initial series (7) we obtain a power series with the same radius of
convergence. The sum of this series will be the second derivative S00(x) of the
sum of the initial series. Such a process can be continued infinitely. Therefore,
the following statement holds.
Corollary (on the infinite differentiability of a power series).
The sum S(x) of power series (7) is an infinitely differentiable function
on the convergence interval; to find its derivative of order m, it is enough to
find the sum of the series obtained from the initial series by differentiating
its terms the required number of times:
S
(m)
(x) =
∞
X
k=0
ckx
k
(m)
=
∞
X
k=m
ckx
k
(m)
.
18. Taylor series
Real analytic functions and their
expansions into Taylor series 3.15B/26:08 (13:22)
Definition.
A function f is called a real analytic function on the interval (x0−R, x0+R)
if it can be expanded on this interval into a convergent power series centered
at x0:
f(x) =
∞
X
k=0
ck(x − x0)k. (1)
From representation (1), using the theorem on differentiation of a power
series and the corollary on infinite differentiability of a power series, we
obtain that the analytic function is infinitely differentiable on the interval
(x0− R, x0+ R).
Let us express the coefficients ckof the series (1) in terms of the values of
the function f and its derivatives.
If we substitute the value x = x0in relation (1), then in this case all terms
vanish except the term for k = 0, therefore relation (1) with x = x0will take
the form
f(x0) = c0.
So, the coefficient c0is equal to the value of the function f at the point x0.
From the theorem on differentiation of a power series it follows that the
derivative of the function f (x) at any point x ∈ (x0−R, x0+R) can be found
by means of the term-by-term differentiation of the power series (1):
f0(x) =
∞
X
k=0
ck(x − x0)
k
0
=
∞
X
k=1
kck(x − x0)
k−1
=
= c1+ 2c2(x − x0) + 3c3(x − x0)2+ .. . (2)
Substituting the value x = x0into relation (2), we obtain the following
equality:
f0(x0) = c1.
18. Taylor series 199
Thus, the coefficient c1is equal to the value of the first derivative of the
function f at the point x0.
Differentiating equality (2), we obtain the relation defining the second
derivative of the function f in the form of a power series:
f00(x) =
∞
X
k=2
k(k − 1)ck(x − x0)
k−2
=
= 2c2+ 3 · 2c3(x − x0) + 4 ·3c4(x − x0)2+ .. . (3)
Let us substitute the value x = x0in relation (3):
f00(x0) = 2c2.
Thus, c2=
f00(x0)
2
.
Now we obtain a representation of the third derivative of the function f
in the form of a power series:
f
000
(x) =
∞
X
k=3
k(k − 1)(k − 2)ck(x − x0)
k−3
=
= 3 ·2c3+ 4 · 3 ·2c4(x − x0) + 5 ·4 · 3c5(x − x0)2+ .. . (4)
Substitute the value x = x0in relation (4):
f
000
(x0) = 3 ·2c3.
In this case, the factorial appears in the denominator of the coefficient c
3
representation: c3=
f
000
(x0)
3!
.
It is easy to verify that a formula of the form
f
(k)
(x0)
k!
remains valid for any
coefficient ck. Moreover, it will also be true for the initial coefficient of the
series, since the function f is considered to be the derivative f
(0)
and the
value 0! is considered equal to 1.
Let us formulate the obtained results as a theorem.
Theorem (on the properties of a real analytic function).
If the function f is a real analytic function on the interval (x0−R, x0+R),
then it is infinitely differentiable on this interval and is expanded on this
interval into the power series with coefficients that are determined by the
values of the function f and its derivatives at the point x0as follows:
ck=
f
(k)
(x0)
k!
, k = 0, 1, 2, . . .
Substituting the found values of the coefficients ckinto relation (1), we get
the equality that holds for all x ∈ (x0− R, x0+ R):
200 M. E. Abramyan. Lectures on integral calculus and series theory
f(x) =
∞
X
k=0
f
(k)
(x0)
k!
(x − x0)k. (5)
The series on the right-hand side of equality (5) is called the Taylor series
of the function f .
In contrast to the expansion of functions considered earlier by Taylor’s
formula (see [1, Ch. 22]), the sum in (5) includes an infinite number of terms
but there is no remainder term.
A special case of the Taylor series is a series centered at the point x0= 0
and converging on the interval (−R, R):
f(x) =
∞
X
k=0
f
(k)
(0)
k!
xk.
Real analytic functions and the property
of infinite differentiability
An example of an infinitely differentiable function
that does not expand into a Taylor series 3.16A/00:00 (15:20)
If a function is expanded into a Taylor series on a certain interval, then,
by virtue of the properties of power series, it is infinitely differentiable on this
interval. The converse is not true.
Consider the function f(x) = e
−
1
x
2
defined on the set R \ {0} and define
it as 0 at zero:
f(x) =
(
e
−
1
x
2
, x 6= 0,
0, x = 0.
This function is infinitely differentiable at any point x 6= 0 as a superposition of infinitely differentiable elementary functions. Let us show that it has
the same property at zero. First of all, we prove its continuity at this point.
To do this, we find the limit of the function f (x) as x → 0:
lim
x→0
f(x) = lim
x→0
e
−
1
x
2
= lim
t→∞
e
−t
2
= 0.
When calculating the limit, we made the variable change t =
1
x
. Note that
the notation t → ∞ for real numbers t means that the parameter t can take
values that infinitely approach both −∞ and +∞. However, in any case, the
value −t2approaches −∞ and so we obtain a limit equal to 0. Thus, the
limit of the function at zero coincides with its value at this point; this means
that the function f is continuous at the point 0.
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