All nested partial sums of a sequence tend to $0$. Is the sequence constant? $S^0:\mathbb N\to \Bbb R$ is a function. For any $m\in \Bbb N$, we define
$$S^m:\Bbb N\to \Bbb R$$
$$S^m(n)=\sum_{k=1}^n S^{m-1}(k)$$
For each $m\in \Bbb N$, we have:
$$\lim_{n \to \infty}S^m(n)=0$$
Can we deduce $S^0\equiv 0?$
 A: No, there are functions such that all your nested partial sums converge to $0$, but that are nonzero :
Suppose you have constructed the first $k(n)$ terms for $S_0$, such that $S_1(k(n)) = S_2(k(n)) = \ldots = S_n(k(n)) = 0$, and let $\varepsilon > 0$.
You want to pick the next $k(n+1)-k(n)$ terms so that the next partial sums stay bounded by $\varepsilon$, and $S_m(k(n+1)) = 0$ for $0 \le m \le n+1$.
This is always possible : in fact you can uniquely choose the next $n+2$ terms such that $S_m(k(n)+n+2) = 0$ for $0 \le m \le n+1$, by solving an invertible linear system of $n+2$ equations. Then, in order to not go over $\varepsilon$ in the intermediate partial sums, find $N$ such that they stay under $N\varepsilon$. Divide those $n+2$ numbers by $N$, and apply them $N$ times to get what you wanted.

You can also have a more explicit construction :
Pick $S_0(0,1) = +1, 0$, so that $S_0(1) = 0$
Then pick $S_0(2+n) = - S_0(n \mod 2)/2$ for $0 \le n < 4$, so that $S_0(5) = S_1(5) = 0$
Then pick $S_0(6+n) = - S_0(n \mod 6)/3$ for $0 \le n < 18$, so that $S_0(23) = S_1(23) = S_2(23) = 0$
Then pick $S_0(24+n) = - S_0(n \mod 24)/4$ for $0 \le n < 96$, so that $S_0(119) = \ldots = S_3(119) = 0$.
And so on.
A: Hmm, it might be that I did not understand the question correctly, but I find the problem interesting so I'll give it a try and please correct me if I'm wrong.     
Here is, what I understand from the definitions of the $S_m$ with an example at $n=4$: 
$$ \begin{eqnarray}
S^1(4) &=& S^0(1)+ S^0(2)+ S^0(3)+ S^0(4)  \\
 &=&    1 S^0(1)+ 1 S^0(2)+ 1 S^0(3)+ 1 S^0(4) \\
S^2(4) &=& S^1(1)+ S^1(2)+ S^1(3)+ S^1(4)  \\
 &=&    4 S^0(1)+ 3 S^0(2)+ 2 S^0(3)+ 1 S^0(4) \\
S^3(4) &=& S^2(1)+ S^2(2)+ S^2(3)+ S^2(4) \\
 &=& 10 S^0(1)+ 6 S^0(2)+ 3 S^0(3)+ 1 S^0(4) \\
\end{eqnarray} $$ 
and
$$ S^m(n) = \sum_{k=1}^n \binom{k+m}{m}S^0(k)$$ where also $ \lim _{n \to \infty}S^m(n) = 0$ is assumed. I hope I've got this correct so far.     
Now in the example, if we subtract $S^1(4)$ from $S^2(4)$ we get $$S^2(4)-S^1(4) = S^2(3)$$ which is generalizable in the obvious way $$S^m(n)-S^{m-1}(n) = S^m(n-1) $$ or 
$$S^m(n)-S^m(n-1) = S^{m-1}(n) $$
But as $S^m(n)$ goes to zero so does $S^m(n-1)$, and also $ \lim_{n \to \infty}S^m(n)-S^{m-1}(n) = 0 $
If that reasoning is correct, then we have also from $m=1$ only that $$ \lim_{n \to \infty}S^0(n) = 0 $$  
Hmm, that's not the final solution I'm afraid. Did your question actually mean that $S^0(n)=0$ for all $n \in \mathbb N$ ?

[Update]: Ok, after that we can write things in matrix-notation. For shortness let's denote $S^0(n)$ as $a_n$ and $S^m(n)$ as $z_{m,n}$ and let us then rewrite the vector/matrix-product for the $n=4$ example:
$$ 
\begin{array} {lll}  
    & * &\left[ \begin{array}  {rrrr} \quad 1  & \quad 4  & \quad 10  & \quad 20\\
                          1 & 3& 6  &10\\
                          1 & 2& 3  & 4\\
                          1 & 1& 1  & 1
        \end{array} \right]
   \\
  \left[ \begin{array} {r} a_1 & a_2 & a_3 & a_4 \end{array} \right]
 & = &\left[ \begin{array} {rrrrr} \ \  z_{1,4} & \  z_{2,4} & \  z_{3,4} &  \ \  z_{4,4} \end{array} \right]
 \end{array}
$$
If this scheme is extended to infinite size ($ n \to \infty$) in the obvious way, the problem says that we expect that for all k we find $z_{k,n} \to 0$      

To divide&conquer the problem, we can LR-factorize the coefficients matrix, let's call it $C_n$ with the index n for the size such that $C_n = L_n \cdot R_n$ with $L_n$ and $R_n$ triangular matrices. We have then the formal definition
$$ \begin{eqnarray} A_n \cdot C_n &=& Z_n \\ A_n \cdot (L_n \cdot R_n) &=& Z_n \\ \end{eqnarray}$$
We get then the interesting effect, that $L_n$ becomes the lower triangular Pascal-matrix and its entries are fixed for any size; so we can look at the left dot-product $ A_n \cdot L_n = B_n $ first and this looks then like
$$ 
\begin{array} {lll}  
    & * &\left[ \begin{array}  {rrrr}
                    \quad 1  &   &  & \\
                          1 & 1&   &\\
                          1 & 2& 1  & \\
                          1 & \  3& \  3  &  \ 1 & \cdots\\
                      \vdots & \vdots & \vdots & \vdots & \ddots 
        \end{array} \right]
   \\
  \left[ \begin{array} {r} a_1 & a_2 & a_3 & a_4 & \cdots \end{array} \right]
 & = &\left[ \begin{array} {rrrrr} \quad  b_0 &  b_1 &  b_2 &  b_3 & \cdots \end{array} \right]
 \end{array}
$$
where if we understand the values $a_n$ as coefficients of a power series in x we have that the $b_0$ can expressed as a function $f(x)$ at $x=1$ by the power series  $$ f(x) = \sum_{k=0}^\infty a_{1+k} x^k  $$and the $b$'s of the following columns are then the derivatives such that
$$ b_k= x^k {f^{(k)}(x)\over k!} \qquad \qquad \text{ at } x=1 $$
Now the right matrix $R$ is upper triangular; and its entries change with the increasing $n$; we get with the $n$ kept indeterminate
$$ R = \left[ \begin{array} {ccccc} 
 1 & n & \binom{n+1}{2} &\binom{n+2}{3} &\binom{n+3}{4}& \cdots \\
 .  & -1 & -n & -\binom{n+1}{2} &-\binom{n+2}{3} & \cdots \\
 . & .  & 1 & n & \binom{n+1}{2} & \cdots \\
 . & . & .  &- 1 & -n &  \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots &\ddots\\
\end{array} \right] $$
and we want, that the matrix-product of $B_n \cdot R_n = Z_n $ in the limit $ n \to \infty$ arrives at the nullvector $ \lim_{n \to \infty}  Z_n = [0,0,0,...]$.     
From the first column of $R$ we see, that $b_0=f(1)=0$ is required to make $z_1=S^1(n)=0$.
From this seems(!) to follow, that for the second column to provide $z_2=0$ also $b_1=0$ is required and from this for all other columns too. However, because n goed to infinity it might be, that there is some more consideration of the limits required - but I'm not yet sure of this.
But given, that this considerations are all correct, then we see, that $f(x)$ must be a function, where all derivatives at $x=1$ become zero - and I know only one function, which is not constant and has still all derivatives at some (single (!)) point $x_0$ equal to zero, this is $g(x)=\exp(-{1 \over x^2})$      at $x_0=0$ .
Caution: if divergent expressions are involved this might look differently - I don't see this yet. 
So, although this small study was a nice experience I still cannot answer this question to one side so far..., sorry...hmmm...
