13
$\begingroup$

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?)

Partial sums of series with exponent 1

$\endgroup$

migrated from math.stackexchange.com Jun 2 '14 at 0:56

This question came from our site for people studying math at any level and professionals in related fields.

  • 2
    $\begingroup$ I really like this question, and it did not receive too much attention on MSE. So I'm migrating it to MO. $\endgroup$ – davidlowryduda Jun 2 '14 at 0:56
7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Gottfried Helms Jun 2 '14 at 9:22
  • $\begingroup$ You're right! In fact, your sum is indeed a little more natural. I'll change the answer above. $\endgroup$ – Lucia Jun 2 '14 at 11:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.