# Absolutely Convergent Series is Convergent

## Theorem

Let $V$ be a Banach space with norm $\norm {\, \cdot \,}$.

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

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

## Proof

That $\ds \sum_{n \mathop = 1}^\infty a_n$ is absolutely convergent means that $\ds \sum_{n \mathop = 1}^\infty \norm {a_n}$ converges in $\R$.

Hence the sequence of partial sums is a Cauchy sequence by Convergent Sequence is Cauchy Sequence.

Now let $\epsilon > 0$.

Let $N \in \N$ such that for all $m, n \in \N$, $m \ge n \ge N$ implies that:

$\ds \sum_{k \mathop = n + 1}^m \norm {a_k} = \size {\sum_{k \mathop = 1}^m \norm {a_k} - \sum_{k \mathop = 1}^n \norm {a_k} } < \epsilon$

This $N$ exists because the sequence is Cauchy.

Now observe that, for $m \ge n \ge N$, one also has:

 $\ds \norm {\sum_{k \mathop = 1}^m a_k - \sum_{k \mathop = 1}^n a_k}$ $=$ $\ds \norm {\sum_{k \mathop = n + 1}^m a_k}$ $\ds$ $\le$ $\ds \sum_{k \mathop = n + 1}^m \norm {a_k}$ Triangle inequality for $\norm {\, \cdot \,}$ $\ds$ $<$ $\ds \epsilon$

It follows that the sequence of partial sums of $\ds \sum_{n \mathop = 1}^\infty a_n$ is Cauchy.

As $V$ is a Banach space, this implies that $\ds \sum_{n \mathop = 1}^\infty a_n$ converges.

$\blacksquare$

## Proof for Real Numbers

By the definition of absolute convergence, we have that $\ds \sum_{n \mathop = 1}^\infty \size {a_n}$ is convergent.

From the Comparison Test we have that $\ds \sum_{n \mathop = 1}^\infty a_n$ converges if and only if:

$\forall n \in \N_{>0}: \size {a_n} \le b_n$

where $\ds \sum_{n \mathop = 1}^\infty b_n$ is convergent.

So substituting $\size {a_n}$ for $b_n$ in the above, the result follows.

$\blacksquare$

## Proof for Complex Numbers

Let $z_n = u_n + i v_n$.

We have that:

 $\ds \cmod {z_n}$ $=$ $\ds \sqrt { {u_n}^2 + {v_n}^2}$ $\ds$ $>$ $\ds \sqrt { {u_n}^2}$ $\ds$ $>$ $\ds \size {u_n}$

and similarly:

$\cmod {z_n} > \size {v_n}$

From the Comparison Test, the series $\ds \sum_{n \mathop = 1}^\infty u_n$ and $\ds \sum_{n \mathop = 1}^\infty v_n$ are absolutely convergent.

From Absolutely Convergent Real Series is Convergent, $\ds \sum_{n \mathop = 1}^\infty u_n$ and $\ds \sum_{n \mathop = 1}^\infty v_n$ are convergent.

By Convergence of Series of Complex Numbers by Real and Imaginary Part, it follows that $\ds \sum_{n \mathop = 1}^\infty z_n$ is convergent.

$\blacksquare$