Math 3210, Fall 2013
Instructor: Thomas Goller
11 December 2013
Learning Celebration #5 Solutions
Example (14 points)
Give an example of each of the following or state that no example exists:
(1) A sequence {an } of real numbers such that lim sup an 6= lim inf an . (2 points)
{( 1)n }
(2) A series of real numbers that converges conditionally. (2 points)
1
X
( 1)k
k
k=1
(3) A series of real numbers that does not converge but which converges absolutely. (2
points)
No example exists
(4) A sequence of functions that converges pointwise but not uniformly on a subset of R (be
sure to specify the subset). (2 points)
nxo
on R
n
(5) A power series centered at 1 with radius of convergence 0. (2 points)
1
X
k!(x 1)k
k=0
(6) A power series centered at 0 with radius of convergence 2. (2 points)
1
X
xk
k=0
2k
(7) A function that is continuous at 0 but not analytic at 0. (2 points)
f (x) = |x|
Computation (6 points)
(8) Compute the Taylor series centered at 0 for the function sin(x). (4 points)
We compute
8
sin(x)
>
>
>
<cos(x)
sin(n) (x) =
>
sin(x)
>
>
:
cos(x)
if
if
if
if
n⌘0
n⌘1
n⌘2
n⌘3
(mod
(mod
(mod
(mod
4);
4);
,
4);
4)
8
0
>
>
>
<1
sin(n) (0) =
>
0
>
>
:
if
if
if
1 if
n⌘0
n⌘1
n⌘2
n⌘3
(mod
(mod
(mod
(mod
4);
4);
.
4);
4)
Math 3210, Fall 2013
Instructor: Thomas Goller
11 December 2013
Thus the Taylor series for sin(x) centered at 0 is
x3 x5
+
3!
5!
x
1
X
x7
x2k+1
+ ··· =
( 1)k
.
7!
(2k
+
1)!
k=0
(9) Show that the Taylor series in (8) converges on R. (2 points)
Since
( 1)k
lim sup
(2k + 1)!
1/(2k+1)
= lim
1
= 0,
(k!)1/k
the radius of convergence is 1, so the Taylor series converges on R.
Proof (20 points)
(10) Let {ak } be a sequence. Assume there is N 2 N such that for all k > N , ak = 0. Prove
1
X
that the series
ak converges using the definition of convergence. (5 points)
k=1
Proof. Let sn =
so
P1
k=1
Pn
k=1
ak denote the partial sums of
|sn
P1
k=1
ak . Given ✏ > 0, for all n > N ,
sN | = |aN +1 + aN +2 + · · · + an | = 0 < ✏,
ak = s N .
1
X
1
(11) Let p > 1. Use the integral test to prove that the series
converges. (5 points)
kp
k=1
Proof. Using the First Fundamental Theorem of Calculus, we compute
✓ p+1
◆
Z 1
Z s
s
1 p+1
1
p
p
x dx = lim
x = lim
=
s!1 1
s!1
p+1
p+1
p 1
1
R1
p+1
since p + 1 < 0 implies
= 0. In particular, 1 x p dx converges, so
P1 thatp lims!1 s
by the integral test, k=1 k converges.
(12) Guess the interval of convergence of the power series
1
X
k=1
as a claim, and prove the claim. (5 points)
Claim. The interval of convergence of the power series
xk
4 + ( 1)k
1
X
k=1
k
, state your guess
xk
4 + ( 1)k
k
is ( 3, 3).
Math 3210, Fall 2013
Instructor: Thomas Goller
11 December 2013
Proof. Since
1
lim sup
(4 + ( 1)k )k
1/k
= lim sup
1
1
= ,
k
4 + ( 1)
3
the radius of convergence is 3. Since the power series is centered at 0, the interval of
convergence is thus ( 3, 3).
4x
. Prove that {fn } converges unin
formly to the constant function 1 on [ 5, 1] using the definition of uniform convergence.
(5 points)
(13) Let {fn } be the sequence of functions fn (x) = 1 +
Proof. Given ✏ > 0, choose N =
|fn (x)
20
.
✏
Then for all n > N and for all x 2 [ 5, 1],
1| = 1 +
4x
n
1 =
4|x|
20
20

<
= ✏.
n
n
N