The ratio and root test also work but the ratios and

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. Arzela-Ascoli 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 Arzela-Ascoli.
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 Arzela-Ascoli, { 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 [