However, it should be clear that
∞
X
k
=1
a
nk
> a
kk
=
k
for all
n
. So we have here a situation where each series
∑
∞
k
=1
a
nk
converges (absolutely since all terms are positive),
the series coming from the limit sequence
{
a
k
}
converges even more so since all terms are 0, but (1) is quite false.
3.
Several of you got 5 for this exercise for using a correct proof incorrectly. It gave me the impression that you were
not quite sure of what you were doing. Before using that proof, a lot of things have to be proved. Here is a sketch of
how it works.
Let
X
be a normed vector space. Because it is a vector space,
∑
n
k
=1
a
k
makes sense for any finite number
a
1
, . . . , a
k
∈
X
.
If
a
n
∈
X
for
n
= 1
,
2
,
3
, . . .
, one can talk now of the series
∑
∞
n
=1
a
n
and say that it converges iff and only if the sequence
{
∑
n
k
=1
a
k
}
∞
n
=1
converges, and write
a
=
∑
∞
n
=1
a
n
iff lim
n
→∞
∑
n
k
=1
a
k
=
a
; i.e., if lim
n
→∞
k
∑
n
k
=1
a
k

a
k
= 0. We say
the series converges absolutely iff the series of nonnegative real terms
∑
∞
n
=1
k
a
n
k
converges. Then one has:
Theorem 1
If
X
is a complete normed space; i.e., a Banach space, absolute convergence implies convergence.
The proof is similar (or identical) to the proof in the complex case, but it requires a proof.
Another result needed for this other proof of completeness, a result that cannot be used so freely, that requires a proof,
is one that is valid in all metric spaces, namely:
Lemma 2
Let
{
p
n
}
be a Cauchy sequence in a metric space
M
.
Then there is a subsequence
{
p
n
k
}
such that
d
(
p
n
k

1
, p
n
k
)
<
2

k
.
The proof is easy, but at our level it is needed.
The main result involved is the converse of Theorem 1, namely:
Theorem 3
If
X
is a normed vector space in which every absolutely convergent sequence converges, then
X
is a Banach
space.
People who used this approach provided what amounts to a moderately incorrect proof of this result.
All in all, I felt there were too many gaps, and several errors, in the proofs i saw, to warrant more points than I gave.
3. Let
{
f
n
}
be a sequence of continuously differentiable functions on [0
,
1] with
f
n
(0) =
f
0
n
(0) and

f
0
n
(
x
)
 ≤
1 for all
x
∈
[0
,
1] and
n
∈
N
. Show that if lim
n
→∞
f
n
(
x
) =
f
(
x
) for all
x
∈
[0
,
1], then
f
is continuous on [0
,
1]. Must the
sequence converge? Must there be a convergent subsequence?