Convergence of alternating harmonic sums - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T07:23:02Z http://mathoverflow.net/feeds/question/51898 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51898/convergence-of-alternating-harmonic-sums Convergence of alternating harmonic sums Wadim Zudilin 2011-01-13T02:01:55Z 2011-01-13T08:00:06Z <p>I owe the idea of asking this question to Max Muller and <a href="http://mathoverflow.net/questions/26035/" rel="nofollow">his curiosity</a>.</p> <p><em>What is the set of $\alpha$ in the interval $0\le\alpha &lt; 1$ for which the alternating sum</em> $$\sum_{n=1}^\infty\frac{(-1)^{n+[n^\alpha]}}n$$ <em>converges</em>? Here $[\ \cdot\ ]$ denotes the integral part of a number.</p> <p>It clearly converges when $\alpha=0$ and <a href="http://mathoverflow.net/questions/26035/evaluation-of-the-following-series-s-1-2-times3-1-5-times6-1-7-tim/48433#48433" rel="nofollow">my post</a> to Max's answer implies the convergence when $\alpha=1/2$. What about more general $\alpha$? Of course, the question is meaningful for positive $\alpha\notin \mathbb Z$ as well, but then it seems to be much harder.</p> http://mathoverflow.net/questions/51898/convergence-of-alternating-harmonic-sums/51922#51922 Answer by Leandro for Convergence of alternating harmonic sums Leandro 2011-01-13T06:16:58Z 2011-01-13T06:16:58Z <p>This is not an answer, but it is too long for a comment.</p> <p>Hi Wadim, nice problem. I was trying to obtain a partial answer for it based on the following </p> <p><b>Proposition.</b> Let be $\xi_1,\xi_2,\ldots$ a sequence of independent Bernoulli random variables with $\mathbb{P}(\xi_n=+1)=\mathbb{P}(\xi_n=-1)=\frac{1}{2}$, then the series $\sum \xi_n a_n$, with $|a_n|\leq c$, converges with probability 1, if and only if $\sum a_n^2&lt;\infty$.</p> <p>The idea it was consider $a_n=\frac{1}{n}$, and try to analyse the set <code>$$A:=\left\{\Big((-1)^{n+[n^{\alpha}]}\Big)_{n\in\mathbb{N}}; 0\leq \alpha&lt;1 \right\} \subset \{-1,1\}^{\mathbb{N}}.$$</code><br> In case that $\mathbb{P}(A)\neq 0$, since this product measure has no atoms we could, at least, say that the set of $\alpha$'s for which the series is convergent is non-enumerable. </p> http://mathoverflow.net/questions/51898/convergence-of-alternating-harmonic-sums/51926#51926 Answer by Anthony Quas for Convergence of alternating harmonic sums Anthony Quas 2011-01-13T06:31:33Z 2011-01-13T06:31:33Z <p>It converges for all $0\le\alpha&lt;1$. Define the $k$-<em>block</em> to be the set of $n$ such that $[n^\alpha]=k$ (it ranges from $\lceil k^{1/\alpha}\rceil$ to $\lceil (k+1)^{1/\alpha}\rceil-1$). </p> <p>The absolute value of the contribution to the sum from the $k$-block is at most the reciprocal of its left endpoint (the terms form an alternating series) - that is approximately $k^{-1/\alpha}$. Hence the contributions from the $k$-blocks are absolutely summable. Bingo</p> http://mathoverflow.net/questions/51898/convergence-of-alternating-harmonic-sums/51931#51931 Answer by Anthony Quas for Convergence of alternating harmonic sums Anthony Quas 2011-01-13T07:54:42Z 2011-01-13T08:00:06Z <p>Let me try and give an answer for $\alpha>1$ also. This one uses some technology from a <a href="http://www.math.uvic.ca/faculty/aquas/paper18.pdf" rel="nofollow">paper</a> of mine with Boshernitzan, Kolesnik and Wierdl.</p> <p>I want to use exponential sums. Using the notation of that paper we take $a(n)=n+n^\alpha$. That paper lets us control $\hat A_t(1/2)=(1/t)\sum_{n\le t}e([a(n)]/2)=(1/t)\sum_{n\le t}(-1)^{n+[n^\alpha]}$ where $e(x)=e^{2\pi i x}$ (see after Lemma 7.2).</p> <p>The proofs of Theorem 3.4 and Theorem 7.1 give (if you read carefully) the existence of an $\epsilon>0$ and a $C$ such that $|\hat A_t(1/2)| &lt; Ct^{-\epsilon}$ for all $t$. This says that the difference between the number of $+1$'s and the number of $-1$'s (the <em>discrepancy</em>) for $n\le t$ is at most $t^{1-\epsilon}$ (ignoring constants from now on). </p> <p>Now let $K>2/\epsilon$ and divide the integers into blocks $I_j=(j^K,(j+1)^K]$. The discrepancy up to $j^K$ is at most $j^{K-2}$ by the above. Similarly the discrepancy up to $(j+1)^K$ is also at most $j^{K-2}$. So the discrepancy in the $I_j$ block is at most $j^{K-2}$. </p> <p>We now have $\sum_{n\in I_j}(-1)^{n+[n^\alpha]}/n = \sum_{n\in I_j}(-1)^{n+[n^\alpha]}/j^K + \sum_{n\in I_j}(-1)^{n+[n^\alpha]}(1/n-1/j^K)$.</p> <p>By the above comment, the first term contributes $j^{-2}$. In the second term, there are $j^{K-1}$ terms, each contributing in absolute value at most $j^{-K-1}$ giving a maximum total contribution of $j^{-2}$.</p> <p>It follows that the contributions from the $I_j$-blocks are absolutely summable and we're done.</p> <p>Of course $\alpha&lt;0$ is trivial so this is good for all $\alpha$ except the positive integers.</p>