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Unformatted text preview: Assuming that the three series
and
and
y and e xy we can interpret Cauchy’s theorem as saying that
and e
e x y e x e y .
x n /n! y n /n! Some Exercises on Cauchy Products
1. Suppose that f n is a sequence of functions defined on the set P of nonnegative integers and that f is a
function defined on P. Suppose that the condition
lim f j f j
nÝn
holds for every nonnegative integer j. Suppose that it is possible to find a sequence n of nonnegative
numbers such that the series n converges and such that
f n j  j for all j and n. Prove that
Ý lim
nÝ Ý fn j
j0 fj.
j0 This exercise is a very slight variation of the proof of the technical lemma that precedes Mertens’
theorem. The solution below is almost a carbon copy of the proof of that lemma.
We remark first that for every j we have f j  j and so the absolute convergence of the series
f j and the series
f n j for any given positive integer n is guaranteed by the comparison test.
In order to show that 347 Ý Ý fn j lim nÝ fj, j0 j0 0. Choose a positive integer N 1 such that suppose that Ý j 3 jN 1 . Now that N 1 has been chosen, choose a positive integer N 2 such that whenever n
N1 N 2 we have N1 fn j fj j0 3 j0 . We define N to be the larger of the two numbers N 1 and N 2 . Now we observe that whenever n
Ý Ý fn j N1 fj j0 j0 Ý N1 Ý fj fn j fn j j0 j0 N1 N1 fj jN 1 1
Ý fn j fj j0 jN 1 1
Ý fn j j0 N we have jN 1 1 fj
jN 1 1 3 3 3 . 2. Prove that if 0 and 1 then the Cauchy product of the series
1n
1n
and
n1
n1
is convergent.
Since both series converge and the second series converges absolutely, the desired result follows
at once from Mertens’ theorem
3. Given that
n Sn
j0 1
j1 for each n, prove that
Ý
n0 1n
S 1 log 2
2
n2 n 2 Hint: Make use of an exercise proved in the document that provides a sharper form of the integral test.
To reach the exercise, click here. If you have not read that integral test document, you can look for the
same exercise when we study power series in this chapter of the text.
4. Prove that if and are positive numbers and 1 then the Cauchy product of the series
1n
1n
and
n1
n1
diverges.
The nth term of this Cauchy product is
n
j0 1nj
n j1 n 1j
j1 1 n
j0 n 1
j1 j1 and since
whenever 0 j n j1 j1
n we see that n1 348 n1 n1 n1 n 1 n 1
j1 j1 n
j0 n 1 1.
n1 j0 Since the nth term of the Cauchy product does not approach 0 as n
diverges. Ý, the Cauchy product 5. Suppose that c n is the Cauchy product of two convergent series a n and b n and suppose that, for
some number 0, the sequences na n log n and n b n are bounded. Prove that the series c n
converges. Hint: Apply Neder’s theorem with n n for each n. Some Exercises on the Cantor Set
1. Prove that the Cantor set does not include any interval of positive length.
Given any positive integer n, an interval that is included in the set E n must have a length not
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.
 Fall '08
 STAFF
 Math, Calculus

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