TITLE: All nested partial sums of a sequence tend to $0$. Is the sequence constant?
QUESTION [14 upvotes]: $S^0:\mathbb N\to \Bbb R$ 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 $S^0\equiv 0?$

REPLY [3 votes]: 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.