8.6 Alternating Series

8.6 Alternating Series - 8 .6 
A l t e r n a t in g 
S...

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Unformatted text preview: 8 .6 
A l t e r n a t in g 
S e r ie s ,
A b s o l u t e 
a n d 
C o n d it io n a l 
 C o n v e rg e n c e 
 ∞ (−1) n +1 an 
 Alternating
series
have
the
form
 ∑ n =1 a1
–
a2
+
a3
–
a4
+
….
 
 Alternating
Series
Test

 ∞ (−1) n +1 an 
converges
if:
 ∑ n =1 1.
an
>
0
for
all
n
 2.
an
≥
an+1

for
all
n
≥
N
for
some
integer
N
 l 3.
 nim an = 0 
 →∞ 
 ∞ Ex.
 
 
 
 
 
 ∑ (−1) n =1 n +1 1 n
 ∞ Ex.
 ∑ (−1) n =1 n +1 2n 2 4 n 2 + 1
 
 
 
 
 
 Note:

The
nth
term
test
always
shows
divergence.

However,
if
 any
of
the
other
criteria
fail
in
the
alternating
series
test,
it
 does
not
mean
that
the
series
diverges.

(More
at
the
end
of
 these
overheads.)
 
 Alternating
Series
Estimation
Theorem
 ∞ (−1) n +1 an 
satisfies
the
three
 If
the
alternating
series
 ∑ n =1 conditions
of
the
Alternating
Series
Test,
then
for
all
n
≥
N
 n sn = ∑ (−1) ai = a1 − a2 + a3 − ... + (−1) i =1 i +1 n +1 an 
 approximates
the
sum
L
of
the
series
with
an
error
whose
 absolute
value
is
less
than
an+1,
the
value
of
the
first
unused
 term.


The
remainder
L
–
sn
has
the
same
sign
as
the
first
 unused
term.
 ∞ Ex.
 ∑ (−1) n =1 n +1 n n 3 + 1
 
 
 
 
 
 
 1.

Estimate
the
error
in
using
the
sum
of
the
first
four
terms
to
 approximate
the
sum
of
the
entire
series.
 
 
 
 2.
Approximate
the
sum
so
the
error
has
magnitude
≤
.01.
 
 
 
 
 
 
 
 Do:

1.
Are
the
following
series
convergent
or
divergent?
 ∞ ∑ (−1) a.
 n +1 n =1 ∞ ∑ c.
 1 10 n 
 ∞ b.
 1 n10 
 n =1 ∑ n =1 ∞ d.
 1 10 n 
 ∑ (−1) n +1 n =1 1 10 − n 
 
 2.

How
many
terms
of
the
series
do
we
need
to
add
in
order
to
 find
the
sum
if
the
error
≤.1?
 ∞ ∑ (−1) n =1 
 
 
 
 
 
 n +1 1 2n − 1 
 Absolute
and
Conditional
Convergence
 ∞ A
series
 ∑ n =1 an 
converges
absolutely
(is
absolutely
 convergent)
if
the
corresponding
series
of
absolute
values,
 ∞ ∞ ∑ an 
,
converges.


If
 ∑ an n =1 ∞ ∑ an 
converges
then
 n =1 
 converges.

(Sometimes
a
series
is
convergent,
but
the
 alternating
series
test
fails.

If
you
can
show
it
is
absolutely
 
 convergent,
then
it
is
convergent.) n =1 ∞ Ex.
 ∑ (−1) n +1 1 n2 
 n +1 1 3 n
 n =1 
 
 ∞ Ex.
 ∑ (−1) n =1 
 
 
 ∞ If
 ∑ ∞ an ∑ 
diverges
but
 n
=1 conditionally
convergent. n =1 an 
converges,
then
the
series
is
 ...
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This note was uploaded on 10/02/2011 for the course AERO 1234 at Virginia Tech.

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