Proof:It only remains to verify the last series converges absolutely. Letting p_{nk} equal 1 if
k ≤ n and 0 if k > n. Then by Theorem 5.4.5 on Page 233
∞ ∞ | n |
∑ |c | = ∑ ||∑ a b ||
n=0 n n=0||k=0 k n−k||
∑∞ ∑n ∑∞ ∑∞
≤ |ak||bn−k| = pnk |ak||bn−k|
n=0k=0 n=0k=0
∑∞ ∑∞ ∑∞ ∑∞
= pnk|ak||bn−k| = |ak||bn−k|
k=0n=0 k=0n=k
∑∞ ∑∞
= |ak| |bn| < ∞. ■
k=0 n=0
The above theorem is about multiplying two series. What if you wanted to consider
( ∞∑ )p
an
n=0
where p is a positive integer maybe larger than 2? Is there a similar theorem to the
above?
Definition 8.11.2Define
∑
ak1ak2 ⋅⋅⋅akp
k1+⋅⋅⋅+kp=m
as follows. Consider all ordered lists of nonnegative integers k_{1},
⋅⋅⋅
,k_{p}which have theproperty that∑_{i=1}^{p}k_{i} = m. For each such list of integers, form the product, a_{k1}a_{k2}
⋅⋅⋅
a_{kp}and then add all these products.
Note that
∑n ∑
akan−k = ak1ak2
k=0 k1+k2=n
Therefore, from the above theorem, if ∑a_{i} converges absolutely, it follows
( )2 ( )
∑∞ ∑∞ ∑
ai = ak1ak2 .
i=0 n=0 k1+k2=n
It turns out a similar theorem holds for replacing 2 with p.
Theorem 8.11.3Suppose∑_{n=0}^{∞}a_{n}converges absolutely. Then if p is a positiveinteger,
( )p
∑∞ ∞∑
an = cmp
n=0 m=0
where
∑
cmp ≡ ak1 ⋅⋅⋅akp.
k1+⋅⋅⋅+kp=m
Proof:First note this is obviously true if p = 1 and is also true if p = 2 from the above
theorem. Now suppose this is true for p and consider
∑ ∞
( n=0an)
^{p+1}. By the induction
hypothesis and the above theorem on the Cauchy product,
(∑∞ )p+1 ( ∞∑ )p (∑∞ )
an = an an
n=0 ( n=0 ) (n=0 )
∑∞ ∞∑
= cmp an
m=(0 n=0)
∑∞ ∑n
= ckpan− k
n=0 k=0
∑∞ ∑n ∑
= ak1 ⋅⋅⋅akpan−k
n=∞0k=0k1+⋅⋅⋅+kp=k
∑ ∑
= ak1 ⋅⋅⋅akp+1■
n=0k1+⋅⋅⋅+kp+1=n
This theorem implies the following corollary for power series.
Corollary 8.11.4Let
∞∑
an(x− a)n
n=0
be a powerseries having radius of convergence, r > 0. Then if
|x − a|
< r,
(∑∞ )p ∞∑
an (x− a)n = bnp(x− a)n
n=0 n=0
where
∑
bnp ≡ ak1 ⋅⋅⋅akp.
k1+⋅⋅⋅+kp=n
Proof: Since
|x− a|
< r, the series, ∑_{n=0}^{∞}a_{n}
(x− a)
^{n}, converges absolutely. Therefore,
the above theorem applies and
( )
∑∞ n p
an (x − a) =
n=0
∞ ( )
∑ ( ∑ k1 kp)
n=0 ak1 (x− a) ⋅⋅⋅akp (x − a) =
k1+⋅⋅⋅+kp=n
( )
∑∞ ∑ n
( ak1 ⋅⋅⋅akp) (x − a) .■
n=0 k1+⋅⋅⋅+kp=n
With this theorem it is possible to consider the question raised in Example 8.8.3 on Page
486 about the existence of the power series for tanx. This question is clearly included in the
more general question of when
( ∞ )− 1
∑ a (x− a)n
n=0 n
has a power series.
Lemma 8.11.5Let f
(x)
= ∑_{n=0}^{∞}a_{n}
(x − a)
^{n}, a power series having radius ofconvergence r > 0. Suppose also that f
(a)
= 1. Then there exists r_{1}> 0 and
{bn}
such thatfor all
|x− a|
< r_{1},
∑∞
--1--= bn(x − a)n .
f (x) n=0
Proof:By continuity, there exists r_{1}> 0 such that if
|x − a|
< r_{1}, then
∑∞
|an||x − a|n < 1.
n=1
Now pick such an x. Then
--1-- = ---∑-∞--1--------n
f (x) 1+ n=1an (x− a)
= ---∑----1---------
1+ ∞n=0cn (x − a)n
where c_{n} = a_{n} if n > 0 and c_{0} = 0. Then
| |
||∑∞ n|| ∞∑ n
|| an (x − a) || ≤ |an ||x− a| < 1 (8.31)
n=1 n=1
(8.31)
and so from the formula for the sum of a geometric series,
by 8.31 and the formula for the sum of a geometric series. Since the series of 8.32 converges
absolutely, Theorem 5.4.5 on Page 233 implies the series in 8.32 equals
( )
∑∞ ∞∑ n
bnp (x − a)
n=0 p=0
and so, letting ∑_{p=0}^{∞}b_{np}≡ b_{n}, this proves the lemma. ■
With this lemma, the following theorem is easy to obtain.
Theorem 8.11.6Let f
(x)
= ∑_{n=0}^{∞}a_{n}
(x− a)
^{n}, a powerseries having radius ofconvergence r > 0. Suppose also that f
(a)
≠0. Then there exists r_{1}> 0 and
{bn}
such that forall
|x− a|
< r_{1},
∑∞
--1--= bn(x − a)n .
f (x) n=0
Proof: Let g
(x)
≡ f
(x)
∕f
(a)
so that g
(x)
satisfies the conditions of the above lemma.
Then by that lemma, there exists r_{1}> 0 and a sequence,
{bn }
such that
f-(a) ∑∞ n
f (x) = bn(x − a)
n=0
for all
|x− a|
< r_{1}. Then
1 ∑∞ n
f-(x) = b^n(x − a)
n=0
where
^bn
= b_{n}∕f
(a)
. ■
There is a very interesting question related to r_{1} in this theorem. Consider f
(x)
= 1 + x^{2}.
In this case r = ∞ but the power series for 1∕f
(x)
converges only if
|x|
< 1. What happens is
this, 1∕f
(x)
will have a power series that will converge for
|x− a|
< r_{1} where r_{1} is the
distance between a and the nearest singularity or zero of f
(x)
in the complex plane. In the
case of f
(x)
= 1 + x^{2} this function has a zero at x = ±i. This is just another instance of why
the natural setting for the study of power series is the complex plane. To read more
on power series, you should see the book by Apostol [3] or any text on complex
variable.