User tambay - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-21T19:52:07Zhttp://mathoverflow.net/feeds/user/21322http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/88209/random-hyperharmonic-seriesrandom hyperharmonic seriestambay2012-02-11T16:23:35Z2012-02-11T18:00:42Z
<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>