# Cesaro(?)/Euler(?) - summation of the $s(p)=\sum_{k=0}^\infty (-1)^{H(k)} (1+k)^p$ for $p=1,2,3,…$ (where $H(k)$ is the Hamming-weight)

In another thread (in MO) there was a question about a series where the signs at the terms alternate with the "Hamming-weight", that means according to the number of bits in the binary representation of the index $k$. That means, for the first few terms of $$S(x)= \sum_{k=0}^\infty (-1)^{H(k)} x^k$$ we have $$S(x)=1-x-x^2+x^3 \quad -x^4+x^5+x^6-x^7 \quad -x^8+x^9+x^{10}-x^{11} \pm ...$$ where the sequence of signs follows the same pattern as the digits in the Thue-Morse-constant or the signs at the powers of $x$ in the expansion of the infinite product$$S(x)=(1-x)(1-x^2)(1-x^4)(1-x^8) ...(1-x^{2^k})...$$

Just for exercising I tried the divergent summation of a type of related series $$s(p)=\sum_{k=0}^\infty (-1)^{H(k)} (1+k)^p \tag 1$$ by which we have $$s(p)=1^p-2^p-3^p+4^p \quad -5^p+6^p+7^p-8^p \quad -9^p+10^p+11^p-12^p \pm ...$$

I tried summation-procedures for divergent summation of $s(1)$, $s(2)$ and $s(3)$ and to my surprise the Euler-summation, which is often "stronger" than Cesaro-sum, does not arrive at conclusive approximations (with even $128$ or $256$ terms of the series) , while Cesaro sum (however higher order) seems to do that, and namely it seems that $$s(1) \underset{\mathfrak C(4)}= 0 \tag 2$$ $$s(2) \underset{\mathfrak C(4)}= 0 \tag 3$$ while $$s(3) \underset{\mathfrak C(?)}= ??? \\ \text{no conclusive result} \tag 4$$ where $\mathfrak C(a)$ means Cesaro-summation to order $a$ (I'm using 128 or 256 terms of the series)

Q1: is the assumption about Cesaro-summation and the result $s(1)=s(2)=0$ correct?
Q2: is Cesaro-summation able to sum higher powers for instance $s(3)$ and to which value?
Q3: is Euler-summation really incapable to give an approximant for such alternating-sign series? (And if: why?) • I really like this question, and it did not receive too much attention on MSE. So I'm migrating it to MO. – davidlowryduda Jun 2 '14 at 0:56

We give two arguments: one that shows rigorously that sufficiently large Cesaro means of $s(p)$ tend to zero for all $p=0$, $1$, $\ldots$, and the other that gives a quick indication of why this is so. Note: I think I got the standard version of Cesaro summation right below, but there are several equivalent forms (in a previous edit I had worked with Riesz means instead of Cesaro means) and in any case the argument would be substantially the same.

First Argument.
Put as in the problem $$S(z) = \prod_{k=0}^{\infty} (1-z^{2^k}) = \sum_{n=0}^{\infty} (-1)^{H(n)} z^n,$$ which is an analytic function in $|z|<1$. Define also for each non-negative integer $k$, $$H_k(z) = \sum_{n=0}^{\infty} \binom{n+k}{k} z^n= \frac{1}{(1-z)^{k+1}}.$$ We are interested in the Cesaro means $$C(N;p,k) = \binom{N+k}{k}^{-1} \sum_{n=0}^{N} (-1)^{H(n)} (n+1)^p \binom{N-n+k}{k},$$ and we wish to show that if $k\ge p+2$ then $C(N;p,k) \to 0$ as $N\to \infty$.

Let $0<r<1$ be a real number to be chosen later. Note that $$\frac{d^p}{d\theta^p} (re^{i\theta} S(re^{i\theta}) ) = \frac{d^p}{d\theta^p} \sum_{n=0}^{\infty} (-1)^{H(n)} (re^{i\theta})^{n+1} = i^p \sum_{n=0}^{\infty} (-1)^{H(n)} (n+1)^p (re^{i\theta})^{n+1}.$$ Therefore by Parseval $$\frac{1}{2\pi} \int_0^{2\pi} \Big(\frac{d^p}{d\theta^p} (re^{i\theta}S(re^{i\theta}))\Big) H_k(re^{i\theta}) (re^{i\theta})^{-N-1} d\theta = i^p \binom{N+k}{k} C(N;p,k).$$ Integrating by parts several times, the LHS above equals $$\frac{(-1)^p}{2\pi } \int_0^{2\pi} (re^{i\theta} S(re^{i\theta}) ) \Big(\frac{d^p}{d\theta^p} (H_k(re^{i\theta}) (re^{i\theta})^{-N-1})\Big) d\theta. \tag{1}$$

Now we use the following bounds for $|S(z)|$ and the derivatives of $H_k(z)$. Note that for any $z$ with $|z|<1$ and any natural number $K$ we have $$|S(z)| = \prod_{k=0}^{\infty} |1-z^{2^{k}}| \le \Big(\prod_{k=0}^{\infty} (1+|z|^{2^k})\Big) \Big(\prod_{k=0}^K (2^k |1-z|)) \Big) \le C_K \frac{|1-z|^{K+1}}{(1-|z|)},$$ for some constant $C_K$. Next note that for any non-negative integer $j$ $$\Big| \frac{d^j}{d\theta^j} H_k(re^{i\theta}) \Big| = \Big|\sum_{n=0}^{\infty} \binom{n+k}{k} n^j r^n e^{-in\theta}\Big| \le D_{k+j} \Big( 1+ \frac{1}{|1-re^{i \theta}|^{k+j+1}}\Big)$$ for some constant $D_{k+j}$. It follows that $$\Big|\frac{d^p}{d\theta^p} (H_k(re^{i\theta}) (re^{i\theta})^{-N-1}) \Big| \le r^{-N-1} A(k,p) \sum_{j=0}^{p} N^{p-j} \Big( 1+\frac{1}{|1-re^{i\theta}|^{k+j+1}} \Big)$$ for some constant $A(k,p)$.

We take $r= 1-1/N$, and use these bounds in (1). Thus we get that this quantity is $$\ll_{K,p,k} \int_0^{2\pi} N |1-re^{i\theta}|^{K+1} N^{p} \Big(1 + \frac{1}{|1-re^{i\theta}|^{k+1} }\Big) d\theta \ll N^{p+1},$$ upon choosing $K \ge k$. We conclude that $$|C(N;p,k)| \ll N^{p+1-k},$$ so that this tends to zero for large $N$ if $k\ge p+2$.

Second Argument. Consider the Dirichlet series $$F(s) = \sum_{n=0}^{\infty} (-1)^{H(n)}(n+1)^{-s},$$ which converges absolutely for Re$(s)>1$. We will obtain a meromorphic continuation for this, which suggests the proper way of renormalizing the sums in the question.

Now consider, following Riemann, $$\int_0^{\infty} e^{-y}S(e^{-y}) y^{s} \frac{dy}{y}.$$ In the region Re$(s)>1$ we may expand the above as $$\int_0^{\infty} \sum_{n=0}^{\infty} (-1)^{H(n)} e^{-(n+1)y} y^s \frac{dy}{y} = \Gamma(s) \sum_{n=0}^{\infty} (-1)^{H(n)} (n+1)^{-s}= \Gamma(s)F(s). \tag{2}$$ Now we examine the LHS above. Since $e^{-y}S(e^{-y})=O(e^{-y})$ as $y\to \infty$ clearly $$\int_{1}^{\infty} e^{-y}S(e^{-y}) y^{s} \frac{dy}{y}$$ is an analytic function for all $s\in {\Bbb C}$. Next note that as $y\to 0$ we have $e^{-y}S(e^{-y}) = O(y^K)$ for any positive integer $K$. Therefore $$\int_0^1 e^{-y}S(e^{-y}) y^s \frac{dy}{y}$$ is also an analytic function of $s$ for all $s\in {\Bbb C}$. We conclude that the LHS of (2) extends analytically to ${\Bbb C}$.

To recap, $\Gamma(s) F(s)$ is analytic in ${\Bbb C}$. Since $\Gamma$ is never zero, and $\Gamma$ has poles at $s=0$, $-1$, $-2$, $\ldots$, we find that $F(s)$ is analytic everywhere, and $F(0)=F(-1)=F(-2)=\ldots =0$.
Thus the regularized values for $s(0)$, $s(1)$, $s(2)$, etc should all be zero.

• Lucia, thanks for this work-out. I'll have to chew on it longer... Just one question: are you sure of the exponent $H(n)+1$ in your first equation (as well in the following ones)? If we want the correct consecutive signs as assumed in the expansion of my series $s(p)$ I think it should be $H(n-1)$ instead. – Gottfried Helms Jun 2 '14 at 9:22
• You're right! In fact, your sum is indeed a little more natural. I'll change the answer above. – Lucia Jun 2 '14 at 11:21