An elementary number theoretic infinite series - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T23:26:55Z http://mathoverflow.net/feeds/question/18483 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18483/an-elementary-number-theoretic-infinite-series An elementary number theoretic infinite series Gil Kalai 2010-03-17T12:18:11Z 2010-11-03T13:57:01Z <p>For a positive integer k, let d(k) be the number of divisors of k. So d(1)=1, d(p) =2 if p is a prime, d(6)=4, and d(12)=6. </p> <p>What is the precise asymptotics of SUM_{k=1}^n 1/(kd(k))</p> <h2>Background:</h2> <p>1) This came up on the side in the <a href="http://gowers.wordpress.com/2010/03/13/edp12-representing-diagonal-maps/#comment-6697" rel="nofollow">polymath5</a> project.</p> <p>2) There, Tim Gowers wrote: If nobody knows the answer, maybe thatâ€™s one for Mathoverflow, where I imagine a few minutes would be enough. </p> <p>3) Asked: 14:17 Jerusalem time. (The first accurate answer: 17:44 Jerusalem time.) </p> <p>4) Looking only at primes or only at integers with a typical number of divisors suggested a loglogn behavior, but looking at semiprimes indicates the sum is larger. I dont know how much larger. </p> <p>5) I couldn't find an asnwer on the web. If there is an easy way searching for an answer that I missed this will be interesting too.</p> <h1>Follow up:</h1> <p>Great answers!! thanks. What about the sum </p> <p>$\sum_{k=1}^n 1/(kd^2(k))$ ?</p> http://mathoverflow.net/questions/18483/an-elementary-number-theoretic-infinite-series/18500#18500 Answer by maks for An elementary number theoretic infinite series maks 2010-03-17T15:44:25Z 2010-03-17T15:44:25Z <p>The correct asymptotic is $C \cdot (\log N)^{1/2}$. (c.f Selberg-Delange method).</p> http://mathoverflow.net/questions/18483/an-elementary-number-theoretic-infinite-series/18501#18501 Answer by David Hansen for An elementary number theoretic infinite series David Hansen 2010-03-17T15:46:35Z 2010-03-18T14:43:59Z <p>(edited)</p> <p>The answer can be extracted from a paper of Ramanujan, "Some formulae in the analytic theory of numbers", no. 17 in his collected papers. There he gives, among other things, the formula</p> <p>$\sum_{n\leq X} \frac{1}{d(n)} \sim \frac{X}{\sqrt{\log{X}}}\pi^{-\frac{1}{2}}\prod_{p}\sqrt{p^2-p}\log{\frac{p}{p-1}}$.</p> <p>The answer to the original question can be extracted from this by partial summation.</p> <p>As for the "follow-up", the answer is $\sum_{n \leq X} \frac{1}{n d(n)^2} \sim C (\log{X})^\frac{1}{4}$. Again, Selberg-Delange...</p> http://mathoverflow.net/questions/18483/an-elementary-number-theoretic-infinite-series/18518#18518 Answer by Victor Miller for An elementary number theoretic infinite series Victor Miller 2010-03-17T20:23:24Z 2010-03-17T20:23:24Z <p>The idea (from the Selberg-Delange) method to doing this problem is the following steps:</p> <p>1) Let $F(s) = \sum_{n\ge 1} \frac{1}{n^s d(n)} = \prod_{p} \left(1 + \sum_{k=1}^{\infty} \frac{1}{(k+1) p^{ks}} \right)$. The latter is by multiplicativity of $d(n)$.</p> <p>2) If we look, instead at $G(s) = \prod_p \left( 1 + \frac{1}{2 p^s} \right)$ we can see that $F(s)/G(s)$ has a non-zero limit as $s \rightarrow 1$ from above. $G(s)$ corresponds in our original sum to restricting $n$ to be square-free.</p> <p>3) $G(s)^2$ almost looks like $\zeta(s)$. Show that $G(s)^2/\zeta(s)$ also has a non-zero limit at $s \rightarrow 1$.</p> <p>4) You then use some Tauberian theorems to show that since $H_n \sim \log n$ (which is the sum associated with $\zeta(s)$ then the corresponding sum for $G(s)$ (i.e. over the square-free $n$) is $\sim to \sqrt{\log n}$.</p>