Vidal mav3385 hw14 schultz 53130 3 converges in the

vidal (mav3385) – HW14 – schultz – (53130) 3
converges. In the first case
a
n
=
parenleftbigg
6 n + 7
4 n
parenrightbigg
n
parenleftbigg
2
3
parenrightbigg
n
=
parenleftbigg
6 n + 7
6 n
parenrightbigg
n
,
in which case
lim
n → ∞
a
n
= e
7 / 6
negationslash = 0;
by the Divergence Test, therefore the series
∞
summationdisplay
n =1
a
n
diverges. On the other hand,
b
n
=
parenleftbigg
6 n + 7
4 n
parenrightbigg
n
parenleftbigg
−
2
3
parenrightbigg
n
= ( − 1)
n
parenleftbigg
6 n + 7
6 n
parenrightbigg
n
= ( − 1)
n
a
n
.
But, as we have seen,
lim
n → ∞
a
n
= e
7 / 6
.
Thus b
n
oscillates between e
7 / 6
and − e
7 / 6
as
n → ∞ ; in particular,
lim
n → ∞
b
n
does not exist. Again by the Divergence Test,
therefore, the series
∞ summationdisplay
n =1
b
n
diverges. Consequently, the given power se-
ries does not converge at x = ± 2 / 3 and so
its
interval of convergence =
parenleftbigg
−
2
3
,
2
3
parenrightbigg
.
005 10.0points
Determine the interval of convergence of
the series
1.
parenleftbigg
−
5
4
, −
1
4
parenrightbigg
correct
2.
parenleftbigg
−
1
4
,
1
4
parenrightbigg
3. ( −∞ , ∞ )
4. series converges only at x = −
3
4
5.
parenleftbigg
1
4
,
5
4
parenrightbigg
Explanation:
The given series has the form
∞
summationdisplay
k =0
c
k
( x − a )
k
with
c
k
= k
3
2
k
, a = −
3
4
.
But then,
lim
k → ∞
c
k +1
c
k
= lim
k → ∞
2
parenleftBig
k + 1
k
parenrightBig
3
= 2 .
By the Ratio Test, the series thus
(i) converges when | x − a | <
1
2
, and
(ii) diverges when | x − a | >
1
2
.
Now at the points x − a =
1
2
and x − a = −
1
2
the series reduces to
∞
summationdisplay
k =0
k
3
,
∞
summationdisplay
k =0
( − 1)
k
k
3
respectively. But in both cases these series di-
verge by the Divergence Test. Consequently,
the interval of convergence of the given series
is
parenleftbigg
a −
1
2
, a +
1
2
.

vidal (mav3385) – HW14 – schultz – (53130) 4
006 10.0points
Determine the interval of convergence of
the infinite series
∞ summationdisplay
k =0
( − 1)
k
( k + 1)!
2
k
(8 x + 6)
k
.
1.
parenleftbigg
− 1 , −
1
2
parenrightbigg
2. series converges only at x = −
3
4
correct
3. ( −∞ , ∞ )
4.
parenleftbigg
1
2
, 1
parenrightbigg
5.
parenleftbigg
−
1
2
,
1
2
parenrightbigg
Explanation:
There are three possibilities for the interval
of convergence of the infinite series
∞
summationdisplay
k =0
a
k
( x − a )
k
.
First set
L = lim
k → ∞
vextendsingle
vextendsingle
vextendsingle
vextendsingle
a
k +1
a
k
vextendsingle
vextendsingle
vextendsingle
vextendsingle
.
Then
(i) when L> 0, the interval of convergence
is given by
parenleftbigg
a −
1
L
, a +
1
L
parenrightbigg
;
(ii) when L = 0, the interval of convergence
is given by
( −∞ , ∞ );
(iii) when L = ∞ , the interval of conver-
gence reduces to the point { a } , i.e. , the series
converges only at x = a .
For the given series,
∞ summationdisplay
k =0
( − 1)
k
( k + 1)!
2 k
(8 x + 6)
k
=
∞
summationdisplay
k =0
( − 1)
k
( k + 1)!8
k
2
k
parenleftbigg
x +
3
4
parenrightbigg
k
,
so the corresponding value of a
k
is
a
k
= ( − 1)
k
8
k
( k + 1)!
2
k
.
In this case
vextendsingle
vextendsingle
vextendsingle
vextendsingle
a
k +1
a
k
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
vextendsingle
vextendsingle
vextendsingle
vextendsingle
−
( k + 2)!8
k +1
2 k +1
·
2
k
( k + 1)!8 k
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
8( k + 2)
2
.