The ratio and root test also work,
But the ratios and roots have to be computed correctly!
. Now
fl
fl
fl
x
n
+1
1+
x
2
n
+2
fl
fl
fl
fl
fl
fl
x
n
1+
x
2
n
fl
fl
fl
=

x

(1 +
x
2
n
)
1 +
x
2
n
+2
.
It is quite obvious that if

x

<
1, then
lim sup
n
→∞

x

(1 +
x
2
n
)
1 +
x
2
n
+2
= lim
n
→∞

x

(1 +
x
2
n
)
1 +
x
2
n
+2
= 0
.
Since 0
<
1, the series converges if

x

<
1.
The limsup is not so easy to calculate if

x

>
1 and I would expect a
Calculus 1 student to have some difficulty with it. But
lim sup
n
→∞

x

(1 +
x
2
n
)
1 +
x
2
n
+2
= lim
n
→∞

x

2
n
+1
(
1 +
1
x
2
n
)
x
2
n
+2
(
1 +
1
x
2
n
+2
)
= lim
n
→∞
1

x

fl
fl
1 +
1
x
2
n
fl
fl
(
1 +
1
x
2
n
+2
)
=
1

x

<
1
.
And we have convergence for

x

>
1. The test fails for

x

= 1, but divergence is obvious (as done before).
The root test also is easy to apply, if one knows what to do with
n
√
1 +
x
2
n
. One probably can’t get away from once
again dividing into the cases

x

>
1,

x

<
1 and

x

= 1. In this last case, the test is inconclusive, so one has to proceed
as before. But it is not incredibly hard to prove that
lim
n
→∞
n
s
fl
fl
fl
fl
x
n
1 +
x
2
n
fl
fl
fl
fl
= lim
n
→∞

x

n
√
1 +
x
2
n
=
0
if

x

<
1,
1
if

x

= 1,
1
/

x

if

x

>
1.
4. Assume
f
n
:
R
→
R
for
n
= 1
,
2
, . . .
are such that
{
f
n
}
is equicontinuous.
(a)
(20 points)
If
{
f
n
}
converges pointwise to some function
f
; i.e., if
f
(
x
) = lim
n
→∞
f
n
(
x
) exists for all
x
∈
R
,
prove
f
is continuous.
Solution.
ArzelaAscoli does 90% of the work, except that Arzela Ascoli is valid only on compact domains.
But continuity is a local property, so no real problem. There is also a quickie proof that bypasses ArzelaAscoli.
Proof.
Let
x
0
∈
R
, we have to prove
f
is continuous at
x
0
. Select any closed and bounded interval such that
x
0
is in its interior; for example, let
a
=
x
0

1
, b
=
x
0
+ 1. The restrictions of the functions
f
n
to the interval [
a, b
]
still form an equicontinuous sequence. If we want to be super perfectly rigorous we can define
g
n
=
f
n

[
a,b
]
; then
obviously
{
g
n
}
is equicontinuous, converges point wise (necessarily) to
f

[
a,b
]
and because it converges pointwise
and is equicontinuous it is also uniformly bounded on [
a, b
] (See below for a proof of this fact, which has already
been proved before in this course). Thus, By ArzelaAscoli,
{
g
n
}
has a uniformly convergent subsequence on [
a, b
],
the limit can only be (once more)
f
.
Thus
f

[
a,b
]
, as the uniform limit of a sequence of continuous functions
(members of an equicontinuous set are trivially continuous) must be continuous on [
a, b
].
Because
x
0
is in the
interior of [