# Absolutely Convergent Series is Convergent iff Normed Vector Space is Banach/Sufficient Condition

## Theorem

Let $\struct {X, \norm {\, \cdot \,}}$ be a normed vector space.

Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be an absolutely convergent series in $X$.

Suppose $\ds \sum_{n \mathop = 1}^\infty a_n$ is convergent.

Then $X$ is a Banach space.

## Proof

Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $X$.

We have that:

$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N: \norm {x_n - x_m} < \epsilon$

We will prove the existence of a subsequence $\sequence {x_{n_k} }_{k \mathop \in \N}$ such that:

$n > n_k \implies \norm {x_n - x_{n_k} } < \dfrac 1 {2^k}$

### Basis for the Induction

Let $\epsilon = \dfrac 1 2$.

Choose $n_1$ such that:

$n, m \ge n_1 \implies \norm {x_n - x_m} < \dfrac 1 2$

In particular, when $m = n_1$:

$n \ge n_1 \implies \norm {x_n - x_{n_1} } < \dfrac 1 2$

### Induction Step

Suppose $x_{n_1}, \dots x_{n_k}$ have been constructed.

Let $\epsilon = \dfrac 1 {2^{k + 1} }$.

Choose $n_{k + 1}$ such that $n_{k + 1} > n_k$ and:

$n, m \ge n_{k + 1} \implies \norm {x_n - x_m} < \dfrac 1 {2^{k + 1} }$

In particular, when $m = n_{k + 1}$:

$n \ge n_{k + 1} \implies \norm {x_n - x_{n_{k + 1} } } < \dfrac 1 {2^{k + 1}}$

$\Box$

Define:

$u_1 := x_{n_1}$
$u_{k + 1} := x_{n_{k + 1} } - x_{n_k}$

Now we have a sequence $\sequence {u_k}_{k \mathop \in \N}$.

Consider the series $\ds \sum_{k \mathop = 1}^\infty \norm {u_k}$:

 $\ds \sum_{k \mathop = 1}^\infty \norm {u_k}$ $=$ $\ds \norm {u_1} + \sum_{k \mathop = 2}^\infty \norm {u_k}$ $\ds$ $=$ $\ds \norm {u_1} + \sum_{k \mathop = 2}^\infty \norm {x_{n_k} - x_{n_{k - 1} } }$ $\ds$ $=$ $\ds \norm {u_1} + \sum_{k \mathop = 1}^\infty \norm {x_{n_{k + 1} } - x_{n_k} }$ Relabeling: $k \to k + 1$ $\ds$ $\le$ $\ds \norm {x_{n_1} } + \sum_{k \mathop = 1}^\infty \frac 1 {2^k}$ $\ds$ $\le$ $\ds \norm {x_{n_1} } + 1$ $\ds$ $<$ $\ds \infty$

Thus, $\ds \sum_{k \mathop = 1}^\infty u_k$ is absolutely convergent.

By assumption in the theorem, $\ds \sum_{k \mathop = 1}^\infty u_k$ is convergent.

In other words:

$\ds \lim_{k \mathop \to \infty} \sum_{j \mathop = 1}^k u_j = u$.

On the other hand:

 $\ds \sum_{j \mathop = 1}^k u_j$ $=$ $\ds x_{n_1} + \sum_{j \mathop = 2}^k \paren {x_{n_j} - x_{n_{j \mathop - 1} } }$ $\ds$ $=$ $\ds x_{n_k}$ Telescoping Series

Therefore:

$\ds \lim_{k \mathop \to \infty} x_{n_k} = u =: x$

So $\sequence {x_{n_k} }_{k \mathop \in \N}$ converges in $X$.

We have that $\sequence {x_{n_k} }_{k \mathop \in \N}$ is a convergent subsequence of a Cauchy sequence$\sequence {x_n}_{n \mathop \in \N}$.

By Convergent Subsequence of Cauchy Sequence in Normed Vector Space, $\sequence {x_n}_{n \mathop \in \N}$ is convergent with the same limit $x$.

By definition, the underlying space is Banach.

$\blacksquare$