Like you say above $$d(a_n,a_m) \leq \sum_{j=n}^{m-1} d(a_j,a_{j+1}) \leq \sum_{j=n}^{m-1} d(a_1,a_2)\cdot k^{j-1} \leq d(a_1,a_2) \sum_{j=n}^{\infty} k^{j-1} = d(a_1,a_2) \frac{k^{n-1}}{1-k}$$
Thus given $\epsilon>0$, choose $N$ such that $d(a_1,a_2)\frac{k^{N-1}}{1-k}< \epsilon$. Then clearly, $m> n \geq N$ imply that $d(a_n,a_m) \leq d(a_1,a_2) \frac{k^{n-1}}{1-k}<\epsilon$.
If I were following the same general route, I would prove by induction on $n$ that
$$0\le\sum_{k=1}^n(-1)^{k+1}a_{m+k}\le a_{m+1}$$
for each $m\in\Bbb Z^+$; since $a_1\ge a_2\ge\ldots\ge 0$, there is no need for absolute value signs.
This is clearly true for $n=1$. Suppose that it is true for some $n\ge 1$; we want to show that
$$0\le\sum_{k=1}^{n+1}(-1)^{k+1}a_{m+k}\le a_{m+1}\tag{0}$$
for any $m\in\Bbb Z^+$. By the induction hypothesis (applied to $m+1$ instead of $m$) we know that
$$0\le\sum_{k=1}^n(-1)^{k+1}a_{(m+1)+k}\le a_{m+2}\;,$$
which can be rewritten as
$$0\le\sum_{k=2}^{n+1}(-1)^ka_{m+k}\le a_{m+2}\;.\tag{1}$$
Now
$$\begin{align*}
\sum_{k=1}^{n+1}(-1)^{k+1}a_{m+k}&=a_{m+1}+\sum_{k=2}^{n+1}(-1)^{k+1}a_{m+k}\\
&=a_{m+1}-\sum_{k=2}^{n+1}(-1)^ka_{m+k}\;,
\end{align*}$$
and it follows from $(1)$ that
$$a_{m+1}-a_{m+2}\le a_{m+1}-\sum_{k=2}^{n+1}(-1)^ka_{m+k}\le a_{m+1}\;.$$
And since $a_{m+1}\ge a_{m+2}$, $(0)$ follows immediately.
Best Answer
$$\forall n\geqslant k,\ \forall m\geqslant k,\ |s_n-s_m|\leqslant a_{k+1}$$