In connection to this question, I found a similar problem in another Miklos Schweitzer contest:

Problem 8./2007 For $A=\{a_i\}_{i=0}^\infty$ a sequence of real numbers, denote by $SA=\{a_0,a_0+a_1,a_0+a_1+a_2,...\}$ the sequence of partial sums of the series $a_0+a_1+a_2+...$. Does there exist a non-identically zero sequence $A$ such that all the sequences $A,SA,SSA,SSSA,...$ are convergent?

If $SA$ is convergent then $A \to 0$. $SSA$ convergent implies $SA \to 0$. We have

  • $SSA=\{a_0,2a_0+a_1,3a_0+2a_1+a_2,4a_0+3a_1+2a_2+a_3...\}$
  • $SSSA=\{a_0,3a_0+a_1,6a_0+3a_1+a_2,10a_0+6a_1+3a_2+a_3...\}$.

I suppose when the number of iteration grows, the coefficients of the sequence grow very large, and I suppose somehow we can get a contradiction if the initial sequence is non-identically 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.

  1. 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 $j=0,1,\ldots,k$.
  2. As $\sum_{n=n_k+1}^\infty 1/n$ is divergent, we can choose $n_{k+1} > n_k$ so that $\sum_{n=n_k+1}^{n_{k+1}}3^{-k}/n > \vert S^{k+1}A_{n_k}\vert$.
  3. 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 $S^{k+1}A_{n_k}+\sum_{n=n_k+1}^{n_{k+1}}b_n=0$.
  4. 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)

  • $\begingroup$ @leonbloy: Why not take $X(z)=\exp((z-1)^{-1})$? This has a zero of infinite multiplicity as you approach $z=1$ from below. So, doesn't your argument say that the terms in the power series expansion of $X(z)$ give a sequence with the required properties? $\endgroup$ – George Lowther Jun 20 '11 at 9:06
  • $\begingroup$ In fact, it gives a sequence such that $\sum_{n=0}^\infty S^kA_nz^n$ has radius of convergence 1 and $\lim_{z\to1}\sum_{n=0}^\infty S^kA_nz^n=0$. It still needs to be shown that you can commute the limit with the summation though. $\endgroup$ – George Lowther Jun 20 '11 at 9:21
  • $\begingroup$ @George : $X(z)=exp((z-1)^{-1}$ is not expressable as a power series in the unit circle, I believe. $\endgroup$ – leonbloy Jun 20 '11 at 12:36
  • $\begingroup$ It is analytic on the unit disc, so it expands as a power series with radius of convergence 1. $\endgroup$ – George Lowther Jun 20 '11 at 12:43
  • $\begingroup$ Yes, I think the problem with my general argument is that the impossibility of a zero of infinite multipliciy applies only for zeros in the interior of the region of convergence. But, for our goal $X(z)$ function, $z=1$ can (must) lie in the border. That is not evident in itself (the border of the ROC is tipically determined by poles, rather than the zeroes), but it's not at all contradictory, and points at the difficulty of constructing the desired sequence. $\endgroup$ – leonbloy Jun 20 '11 at 12:53

I would suggest you try using the alternating harmonic series. It is conditionally convergent so you can try rearrangements that might pop out convergent to zero.


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