Proving this for arbitrary ordered fields is a little
trickier than for Archimedean fields, partly because
there are no concrete sequences -- other than eventually
constant ones -- that are guaranteed to converge or
even to be Cauchy sequences and partly because we lack
the embedding in a completely ordered field.
These problems can be overcome by constructing all the
necessary sequences and series from the one Cauchy
sequence we assume to exist. To begin with, note two
important facts. First, for a Cauchy sequence to
converge, it is sufficient that some subsequence
converges. Second, any sequence has a strictly increasing
subsequence, a strictly decreasing subsequence or a
constant subsequence. For this problem, the latter
case is trivial and the first two can be reduced to each
other by negation, so we need to prove only one of them.
Let $K$ be an ordered field in which every absolutely
convergent series is convergent. If $\{a_n\}$ is a strictly
increasing Cauchy sequence in $K$, then $\{a_n\}$ converges.
Proof:
Let $b_n = a_{n+1} - a_n$. Then ${b_n}$ is positive and
converges to $0$, so it has a strictly decreasing
subsequence $\{b_{n_k}\}$.
Let $c_k = b_{n_k} - b_{n_{k+1}}$. We now have a convergent
series with positive terms $\sum_{k=1}^\infty c_k = b_{n_1}$.
As $\{a_n\}$ is a Cauchy sequence, it has a subsequence $\{a_{m_k}\}$
such that $a_{m_{k+1}} - a_{m_k} < c_k$ for all $k$.
Now consider the series $\sum_{i=1}^\infty d_i$ where
$d_{2k-1} = a_{m_{k+1}} - a_{m_k}$ and
$d_{2k} = a_{m_{k+1}} - a_{m_k} - c_k$.
Note that $-c_k < d_{2k} < 0 < d_{2k-1} < c_k$, so we can pair off
terms to get
$$
\sum_{i=1}^\infty |d_i| = \sum_{k=1}^\infty (d_{2k-1} - d_{2k})
= \sum_{k=1}^\infty c_k = b_{n_1}
$$
By the hypothesis on $K$ we may conclude that $\sum_{i=1}^\infty d_i$
converges and
$$
\sum_{i=1}^\infty d_i + \sum_{k=1}^\infty c_k
= \sum_{k=1}^\infty (d_{2k-1} + d_{2k} + c_k)
= 2 \sum_{k=1}^\infty (a_{m_{k+1}} - a_{m_k}).
$$
Because a Cauchy sequence with a convergent subsequence converges,
we have
$$
\lim_{n \to \infty} a_n = a_{m_1} + \sum_{k=1}^\infty (a_{m_{k+1}} - a_{m_k})
= a_{m_1} + \frac{1}{2}\left(b_{n_1} + \sum_{i=1}^\infty d_i \right)
$$
To the question "how did I come up with this?": there are
not many things that could possibly work. The problem is
set in an environment where none of the power tools of
analysis work. Basic arithmetic works, inequalities work,
some elementary properties of sequences and series work, but
if you want to take a limit of something it'd better be
convergent by hypothesis or by construction.
One more or less obvious attack is by contraposition: assume
that there is a divergent Cauchy sequence and try to construct
a divergent, absolutely convergent series. Such a series
must be decomposable into a positive part $a$ and a negative
part $b$, where $a+b$ diverges and $a-b$ converges. This
is possible in several ways by taking $a$ and $b$ to be linear
combinations of known convergent and divergent series.
A complication is that the terms of the convergent series
must dominate those of the divergent series, as they must
control the signs. I wasted a lot of time trying to get the
convergent series to do this, which is very hard, perhaps
impossible. Then I turned to the proof for vector spaces
for inspiration, and saw that it was in fact very easy to
adjust the divergent series instead, as the partial sums
are a Cauchy sequence. I also adopted the
overall structure of that proof, which is why the final
version is not by contraposition.
The convergence is absolute, provided that $0\leq x<1$. In other words, $x=1$ is not allowed (and Abel's theorem attempts to assign a sensible value to the sum at $x=1$). Since $\sum_{n=0}^\infty a_n$ converges, then $a_n\rightarrow 0$, which means that $|a_n|\leq C$ and so $\sum_{n\geq 0} |a_nx^n|\leq C \sum_{n\geq 0} |x|^n<\infty$ for $0\leq x<1$.
Best Answer
Let $\tau$ be a topology on $\mathbb{R}$ under which every absolutely convergent series converges.
Proof. Let $(y_n)$ be any subsequence of $(x_n)$. Then there exist $(n_k)$ such that $|y_{n_k} - L| < 2^{-k}$. Now consider the sequence $(a_k)_{k=0}^{\infty}$ defined by
$$ a_k = \begin{cases} L, & k = 0 \\ y_{n_j} - L, & k = 2j-1 \text{ for some } j \geq 1 \\ L - y_{n_j}, & k = 2j \text{ for some } j \geq 1 \end{cases} $$
Then $\sum_{k=0}^{\infty} |a_k| < \infty$, and so its partial sum $s_k = \sum_{j=0}^{k} a_j$ converges under $\tau$. Since $s_{2k} = L$, it follows that $s_k \to L$ under $\tau$. Then, since $s_{2k-1} = y_{n_k}$, we have $y_{n_k} \to L$ under $\tau$.
So far, we have proved that every subsequence of $(x_n)$ has a further subsequence that converges to $L$ under $\tau$. This suffices to guarantee that $(x_n)$ converges to $L$ under $\tau$. ////
By this lemma, any series which converges conditionally under the standard topology also converges under $\tau$.