nbn, show the odd partial sums are all no
larger than ∑n=1∞
(− 1)
nbn and are increasing while the even partial sums are at least
as large as ∑n=1∞
(− 1)
nbn and are decreasing. Use this to give another proof of
the alternating series test. If you have trouble, see most standard calculus
books.
Use Theorem 5.3.4 in the following alternating series to tell how large n must be so that
| |
||∑ ∞k=1(− 1)kak − ∑nk=1 (− 1)kak||
is no larger than the given number.
∑k=1∞
(− 1)
k
1k
,.001
∑k=1∞
(− 1)
k
12
k
,.001
∑k=1∞
(− 1)
k−1
1
√k-
,.001
Consider the series ∑k=0∞
(− 1)
n
--1--
√n+1
. Show this series converges and so it makes
sense to write
(∑ )
∞k=0 (− 1)n√n1+1-
2. What about the Cauchy product of this series?
Does it even converge? What does this mean about using algebra on infinite sums as
though they were finite sums?
Verify Theorem 5.4.8 on the two series ∑k=0∞2−k and ∑k=0∞3−k.
You can define infinite series of complex numbers in exactly the same way as infinite
series of real numbers. That is w = ∑k=1∞zk means: For every ε > 0 there exists N
such that if n ≥ N, then
∑
|w − nk=1 zk|
< ε. Here the absolute value is the one which
applies to complex numbers. That is,
|a + ib|
=
√ ------
a2 + b2
. Show that if
{an}
is a
decreasing sequence of nonnegative numbers with the property that limn→∞an = 0
and if ω is any complex number which is not equal to 1 but which satisfies
|ω|
= 1, then ∑n=1∞ωnan must converge. Note a sequence of complex numbers,
{an + ibn}
converges to a + ib if and only if an→ a and bn→ b. See Problem 6
on Page 138. There are quite a few things in this problem you should think
about.