random hyperharmonic series - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T01:18:38Z http://mathoverflow.net/feeds/question/88209 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88209/random-hyperharmonic-series random hyperharmonic series tambay 2012-02-11T16:23:35Z 2012-02-11T18:00:42Z <p>The Harmonic Series is defined as: $\sum_{n} \frac{1}{n}$ where $n=1,2,3,4....$. This series is known to be divergent.</p> <p>A generalization of this series can be made by raising each term to $p$: $\sum_{n} \frac{1}{n^p}$ which is also known as the hyperharmonic series and is known to be convergent when $p>1$.</p> <p>On the other hand, for $p=1$, if the signs of the terms are alternating the sum: $\sum_{n} \frac{(-1)^n}{n}$ is convergent and approaches $\ln{2}$.</p> <p>A natural extension would be to introduce randomness in the sign of each terms. $\sum_{n} \frac{\epsilon_{n}}{n}$ where $\epsilon_{n}$ is defined by the probability of its outcome: $P(\epsilon_{j} = 1)=P(\epsilon_{j} = -1)=1/2$. This is called the random harmonic series in Schmuland (http://www.stat.ualberta.ca/people/schmu/preprints/rhs.pdf).</p> <p>My question is: What would happen if we generalize this to the case of the random HYPERharmonic series? What would be the distribution of the result of the summation?</p> http://mathoverflow.net/questions/88209/random-hyperharmonic-series/88212#88212 Answer by Noam D. Elkies for random hyperharmonic series Noam D. Elkies 2012-02-11T18:00:42Z 2012-02-11T18:00:42Z <p>The case $p=2$ is treated briefly in the final section of the cited Schmuland paper, which gives a picture of the distribution. The observations in the paper's first few sections adapt to arbitrary $p>1$: the distribution is not expected to have a simple formula, but the moment-generating function has a product formula (exhibited below), which we can use to compute each power moment ${\mathbb E}(X_p^{2k})$ of this random variable $X_p = \sum_n \epsilon_n/n^p$ as a polynomial in $\zeta(2p), \zeta(4p), \zeta(6p), \ldots, \zeta(2kp)$. [The odd-order moments vanish by symmetry.] This is because the distribution of $X_p$ is the convolution of an infinite series of distributions, the $n$-th of which is supported on $\pm 1 /n^p$ each with probability $1/2$; therefore $${\mathbb E}(\exp(tX_p)) = \prod_{n=1}^\infty \left(\frac12 e^{t/n^p} + \frac12 e^{-t/n^p} \right) = \prod_{n=1}^\infty \cosh(t/n^p).$$ We can recover the power moments from the expansion of ${\mathbb E}(\exp(tX_p))$ in a Taylor series about $t=0$, writing $$\begin{eqnarray} \log {\mathbb E}(\exp(tX_p)) &amp;=&amp; \sum_{n=1}^\infty \log(\cosh(t/n^p)) \cr &amp;=&amp; \sum_{n=1}^\infty \frac12 \left(\frac{t}{n^p_{\phantom1}}\right)^2 - \frac1{12} \left(\frac{t}{n^p_{\phantom1}}\right)^4 + \frac1{45} \left(\frac{t}{n^p_{\phantom1}}\right)^6 - \frac{17}{2520} \left(\frac{t}{n^p_{\phantom1}}\right)^8 + - \cdots \cr &amp;=&amp; \frac{\zeta(2p)}{2} t^2 - \frac{\zeta(4p)}{12} t^4 + \frac{\zeta(6p)}{45} t^6 - \frac{17\zeta(6p)}{2520} t^8 + - \cdots \end{eqnarray}$$ and exponentiating.</p> <p>While the random variables $X_p$ may be no more than objects of curiosity, similar constructions arise naturally in analytic number theory, such as the value at a fixed $s>1$ of the Dirichlet $L$-series associated to a random real character $\chi$. In this example, the terms $\chi(n)/n^s$ in the Dirichlet series are correlated, but the Euler product $L(s,\chi) = \prod_l (1 - \chi(l)/l^s)^{-1}$ has independent factors, so $\log L(s,\chi)$ is an infinite convolution of the same kind. Likewise if the $\epsilon_n$ were random <em>complex</em> numbers of unit length: the analogue could be $\log L(s,\chi)$ for a random Dirichlet charater $\chi$ that need not be real, or $\log \zeta(\sigma + it)$ for fixed $\sigma>1$ and random real $t$. For these sums of complex-valued random variables, the factors in the moment generating function get more complicated than hyperbolic cosines, but are still tractable.</p>