With thanks to Thomas and Robert:
Suppose $ \sum\limits_{n=1}^\infty a_n$ converges.
Choose a positive number $\alpha$ so that $\tag{1}\left|\,\sum\limits_{n=1}^m a_n\,\right|\le \alpha $ for all positive integers $m$.
Set $S_1=\sum_{n=1}^\infty\, \alpha a_n.$
$S_1$ is convergent, so we may, and do, choose $n_1$ so that for all $l\ge m\ge n_1$ $ \tag{2} \Bigl|\,\sum_{n=m}^l a_i\,\Bigr|\le {1\over 2^2(\alpha+1)} . $
Set $S_2=\underbrace{\sum_{n=1}^{n_1-1} \alpha a_n}_{D_1} + \sum_{n=n_1}^{\infty} (\alpha+1) a_n .$ Note that by (1), $ \left| \, \sum\limits_{n=1}^{m} \alpha a_n\,\right| \le \alpha $ for all $m\le n_1-1$.
Now choose $n_2>n_1$ so that for all $l\ge m\ge n_2$ $ \left|\,\sum_{n=m}^l a_i\,\right|\le {1\over 2^3(\alpha+2)} . $
Set $S_3=\sum_{n=1}^{n_1-1} \alpha a_n + \underbrace{\sum_{n=n_1}^{n_2-1} (\alpha+1) a_n }_{D_2} + \sum_{n=n_2}^{\infty} (\alpha+2) a_n .$
Note that, by (2), $\Bigl|\,\sum\limits_{n=n_1}^{m} (\alpha+1) a_n \,\Bigr|\le {1\over 2^2} $ for all $m\le n_2-1$.
Continuing in the obvious manner, we define integers $ n_3 and sums $D_k=\sum\limits_{n=n_{k-1}}^{n_k-1} (\alpha+k-1)a_n$ satisfying $ \tag{4}\left|\,\sum_{n= n_{k-1}}^{ m}(\alpha+k-1)a_n\,\right|\le {1\over 2^k} $ for all $m\le n_k-1$.
Consider the sum $ S=D_1+D_2+D_3+\cdots. $ We have, by the triangle inequality, that $\eqalign{ |D_n+D_{n+1}+\cdots+ D_m|&\le {1\over 2^n} +{1\over 2^{n+1}} +\cdots+{1\over 2^m} \cr &\le {1\over 2^{n-1} } \cr &\buildrel{n \rightarrow\infty}\over{\longrightarrow }\ 0,} $ for all for $m\ge n>1$.
From this and (4), it follows that $S$ converges.