Many convergent sequences imply the initial sequence zero?
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.
Non rigourous:
Thinking of A as a signal $x[n]$, and using the Z transform, the succesive sums can be thought as $y_k[n]$, the output of cascaded LTI one-pole filters. The resulting filter response is given by $$H_k(z)= \left( \frac{1}{1-z^{-1}} \right)^k$$
with $$Y_k(z) = H_k(z) X(z)$$
We require each $y_k[n]$ to be convergent -and summable-, hence $$Y_k(1) < \infty$$ for all $k$. This would require $X(z)$ to have an zero of infinite multiplicity at $z=1$. Hence, the only possible solution seems to be the identically zero signal (what remains here is to consider if some $x[n]$ can provide us a $X(z)$ with such a infinutely-multiple zero - I'd say no, but I'm not sure).
Update: I think the rigourous proof would involve showing that $X(z)$ is analytic (it's defined as a power series) and hence it may only have zeroes of finite multiplicity (eg)
Update 2: For the benefit of those not familiar with the Z tranforms, here's a brief explanation - I use here $z$ instead of $z^{-1}$. Let
$$X(z) = \sum_{n=0}^\infty x[n] z^n$$
where $x[n]$ is the original sequence and $z$ a complex variable. The series converges in a circle that includes $z=1$, hence $X(z)$ is analytic in that region. Further, let $Y_k(z)$ be the analogous function defined for the $k$ sum. As $y_1[n]=y_1[n-1]+x[n]$ we get $Y_1(z) - z Y_1(z) = X(z) $, hence
$$Y_1(z) = \frac{X(z)}{1-z}$$
And also
$$Y_k(z) = \frac{X(z)}{(1-z)^k}$$
And because all summations must be summable, $Y_k(1) <\infty$, what would require $X(z)$ to have a zero of infinity multiplicity at $z=1$. But then all its derivatives would be zero, and then (as it's analytic) it would be the constant zero function.
(Of course, if George Lowther's answer is right -as it seems to me- this must be flawed; but I don't know where)