Absolutely Convergent Series is Convergent
Contents |
Theorem
Let $V$ be a Banach space with norm $\left\Vert{\cdot}\right\Vert$.
Let $\displaystyle \sum_{n=1}^\infty a_n$ be an absolutely convergent series in $V$.
Then $\displaystyle \sum_{n=1}^\infty a_n$ is convergent.
Proof
That $\displaystyle \sum_{n=1}^\infty a_n$ is absolutely convergent means that $\displaystyle \sum_{n=1}^\infty \left\Vert{a_n}\right\Vert$ 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:
- $\displaystyle \sum_{k=n+1}^m \left\Vert{a_k}\right\Vert = \left\vert{\sum_{k=1}^m \left\Vert{a_k}\right\Vert - \sum_{k=1}^n \left\Vert{a_k}\right\Vert }\right\vert < \epsilon$
This $N$ exists because the sequence is Cauchy.
Now observe that, for $m \ge n \ge N$, one also has:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \left\Vert{\sum_{k=1}^m a_k - \sum_{k=1}^n a_k}\right\Vert\) | \(=\) | \(\displaystyle \left\Vert{\sum_{k=n+1}^m a_k}\right\Vert\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\le\) | \(\displaystyle \sum_{k=n+1}^m \left\Vert{a_k}\right\Vert\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Triangle inequality for $\left\Vert{\cdot}\right\Vert$ | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(<\) | \(\displaystyle \epsilon\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
It follows that the sequence of partial sums of $\displaystyle \sum_{n=1}^\infty a_n$ is Cauchy.
As $V$ is a Banach space, this implies that $\displaystyle \sum_{n=1}^\infty a_n$ converges.
$\blacksquare$
Proof for Real Numbers
By the definition of absolute convergence, we have that $\displaystyle \sum_{n=1}^\infty \left|{a_n}\right|$ is convergent.
From the Comparison Test we have that $\displaystyle \sum_{n=1}^\infty a_n$ converges if:
- $\forall n \in \N^*: \left|{a_n}\right| \le b_n$ where $\displaystyle \sum_{n=1}^\infty b_n$
is convergent.
So substituting $\left|{a_n}\right|$ for $b_n$ in the above, the result follows.
$\blacksquare$
