Absolutely Convergent Series is Convergent

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Theorem

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

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


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


Proof

That $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ is absolutely convergent means that $\displaystyle \sum_{n \mathop = 1}^\infty \norm {a_n}$ 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:

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

This $N$ exists because the sequence is Cauchy.


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

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

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

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

$\blacksquare$


Proof for Real Numbers

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

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

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

where $\displaystyle \sum_{n \mathop = 1}^\infty b_n$ 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:

\(\displaystyle \cmod {z_n}\) \(=\) \(\displaystyle \sqrt { {u_n}^2 + {v_n}^2}\)
\(\displaystyle \) \(>\) \(\displaystyle \sqrt { {u_n}^2}\)
\(\displaystyle \) \(>\) \(\displaystyle \size {u_n}\)

and similarly:

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

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

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

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

$\blacksquare$


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