There do exist non-identically zero sequences with the required properties.
I don't know of any simple description, but can outline a construction which will generate such sequences. First, let me define a bit of notation. If $A=(a_0,a_1,a_2,\ldots)$ then set $DA = (a_0,a_1-a_0,a_2-a_1,\ldots)$, so that $D$ and $S$ are inverses of each other.
Writing $S^kA_n$ for the n'th element of the sequence $S^kA$, I will construct a nonzero series such that, for each $k$, $\vert S^kA_n\vert\le3^{-k}/n$ for all large enough $n$. So, $S^kA$ converges to zero.
We can construct $A$ by building up the initial segments of the sequence in an inductive fashion. Consider the following steps.
- Suppose that, for some integer $k\ge0$ and $n_k > 0$, we have chosen the initial segment $a_0,a_1,\ldots,a_{n_k}$ with the property that $\vert S^jA_{n_k}\vert\le3^{-j}/n_k$ for $j=0,1,\ldots,k$.
- As $\sum_{n=n_k+1}^\infty 1/n$ is divergent, we can choose $n_{k+1} > n_k$ so that $\sum_{n=n_k+1}^{n_{k+1}}3^{-k}/n > \vert S^{k+1}A_{n_k}\vert$.
- Choose real numbers $b_{n_k+1},b_{n_k+2},\ldots,b_{n_{k+1}}$ with $\vert b_n\vert\le 3^{-k}/n$ and $S^{k+1}A_{n_k}+\sum_{n=n_k+1}^{n_{k+1}}b_n=0$.
- Define $a_{n_k+1},a_{n_k+2},\ldots,a_{n_{k+1}}$ so that $S^kA_n=b_n$ for $n_k < n\le n_{k+1}$. (Note: Applying $D^k$ allows us to express $a_n$ in terms of the sequence $S^kA$, so we obtain $a_n$ as a linear combination of the $b_n$).
This construction is designed so that we obtain $S^{k+1}A_{n_{k+1}}=0$ and, in particular, $\vert S^{k+1}A_{n_{k+1}}\vert\le 3^{-k-1}/n_{k+1}$. We also have $\vert S^kA_n\vert\le 3^{-k}/n$ for $n_k\le n\le n_{k+1}$ so, for $n > n_k$, this gives $ \begin{align} \vert S^{k-1}A_n\vert&=\vert S^kA_n- S^kA_{n-1}\vert\\ &\le 3^{-k}/n+3^{-k}/(n-1)\\ &\le 3^{-k}/n+3^{-k}2/n\\ &=3^{-(k-1)}/n. \end{align} $ Applying this inequality iteratively with $k$ replaced by $k-1,k-2,\ldots$ shows that $\vert S^jA_n\vert\le 3^{-j}/n$ for all $j\le k$ and $n_k\le n\le n_{k+1}$. So, we can repeat the above steps with $k$ replaced by $k+1$, then by $k+2$, and so on. It is also possible to initialize the procedure for $k=0$ by taking $n_k=1$ and $a_0=1,a_1=0$.
This leads to a sequence $A=(a_0,a_1,a_2,\ldots)$ and $n_1 < n_2 < n_3 < \cdots$ such that $\vert S^k A_n\vert\le 3^{-k}/n$ whenever $n\ge n_k$. So, for each $k$, $S^kA$ is a sequence converging to 0.
I'll also add an argument showing that, if we expand $f(z)\equiv\exp(1/(z-1))$ as a power series, $ f(z)=\sum_{n=0}^\infty a_nz^n $ then $A=\{a_0,a_1,a_2,\ldots\}$ satisfies the required properties. I can't take full credit for this argument, as it is just following through on the ideas in leonbloy's answer.
The idea is that $f(z)$ is analytic on the disc $\vert z\vert < 1$, so the series expansion is well-defined with radius of convergence equal to 1. Furthermore, $f(z)$ has a zero of infinite order as $z$ approaches 1. More precisely, $(z-1)^{-k}f(z)$ extends to a smooth function on the unit circle $\{\vert z\vert=1\}$. Integrating around the contour $\gamma(t)=\exp(2\pi it)$, Cauchy's integral formula gives $ a_n=\frac{1}{2\pi i}\int_\gamma z^{-n-1}f(z)\,dz=\int_0^1\exp(-2\pi int)f(\gamma(t))\,dt. $ So, $a_n$ is the Fourier series of the smooth function $f(\gamma(t))$ and, therefore, $a_n\to0$ as $n\to\infty$.
Finally, for $k\ge0$, $S^kA$ is the sequence of coefficients in the power series expansion of $(1-z)^{-k}f(z)$. So, applying the same argument as above to $(1-z)^{-k}f(z)$ shows that $S^kA$ converges to 0.