Sum of the reciprocal of perfect numbers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T19:47:25Z http://mathoverflow.net/feeds/question/99227 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99227/sum-of-the-reciprocal-of-perfect-numbers Sum of the reciprocal of perfect numbers Ricky 2012-06-10T10:25:00Z 2012-06-23T20:22:08Z <p>It is well known that the sum of the reciprocal of prime numbers is $+\infty$. This proves that there infinitely many prime numbers. On the other hand it is also known that the series of the reciprocal of twin prime numbers converges, so nothing can be said about the finiteness of the set of twin prime numbers. This is indeed an open problem. A similar open problem is the existence of infinitely many perfect numbers. So this is the question:</p> <p>let $\mathcal{Pe}$ be the set of perfect numbers. What is known about the series $$\sum_{n \in \mathcal{Pe}} \frac{1}{n}$$</p> <p>This is just curiosity, and could be a trivial question. Note that if there are no odd perfect numbers (as it seems to be the case), the series converges. Indeed any $n$ perfect is of the form $2^{p-1}(2^p-1)$.</p> http://mathoverflow.net/questions/99227/sum-of-the-reciprocal-of-perfect-numbers/99229#99229 Answer by Douglas Zare for Sum of the reciprocal of perfect numbers Douglas Zare 2012-06-10T10:56:45Z 2012-06-10T10:56:45Z <p>A Google search finds papers <a href="http://arxiv.org/PS_cache/arxiv/pdf/1101/1101.0259v1.pdf" rel="nofollow">estimating the sum of the reciprocals of amicable numbers</a> and <a href="http://tech.groups.yahoo.com/group/primenumbers/message/23021" rel="nofollow">this discussion</a>:</p> <pre><code>The asymptotic number of perfect numbers less than x was proven in MR0090600 (19,837d) Hornfeck, Bernhard; Wirsing, Eduard Über die Häufigkeit vollkommener Zahlen. (German) Math. Ann. 133 (1957), 431-438. to be less than x^eps for all eps&gt;0. From this it trivially follows that the sum of the reciprocals of the perfect numbers is finite. </code></pre> http://mathoverflow.net/questions/99227/sum-of-the-reciprocal-of-perfect-numbers/99246#99246 Answer by Péter Komjáth for Sum of the reciprocal of perfect numbers Péter Komjáth 2012-06-10T16:48:51Z 2012-06-10T16:48:51Z <p>I think that Erdos proved in 1934 that the sum of the reciprocals of primitive abundant numbers is convergent. A number $n$ is abundant if $\sigma(n)\geq 2n$ and is primitive aboundant if it is aboundant but none of its proper divisor is. It is easy to see that perfect numbers are primitive aboundant and so is the result. P. Erdõs: On the density of the abundant numbers, J. London Math. Soc. 9 (1934), 278--282. Available as <a href="http://www.renyi.hu/~p_erdos/1934-04.pdf" rel="nofollow">http://www.renyi.hu/~p_erdos/1934-04.pdf</a> . </p> http://mathoverflow.net/questions/99227/sum-of-the-reciprocal-of-perfect-numbers/100482#100482 Answer by Péter Komjáth for Sum of the reciprocal of perfect numbers Péter Komjáth 2012-06-23T20:22:08Z 2012-06-23T20:22:08Z <p>There is, indeed a simple argument to show that the sum of the reciprocals of perfect numbers converges. It suffices to show that the number of perfect numbers up to $x$ is $O(\sqrt{x})$. As the number of even perfect numbers is less than $\log x$, it is enough to consider odd perfect numbers. Assume that $n\leq x$ is an odd perfetc number. If $n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_r^{\alpha_r}$, then $2n=(p_1^{\alpha_1}+p_1^{\alpha_1-1}+\cdots+1)(p_2^{\alpha_2}+p_2^{\alpha_2-1}+\cdots+1)\cdots(p_r^{\alpha_r}+p_r^{\alpha_r-1}+\cdots+1)$. As $n$ is odd, exactly one factor on the right hand side is even. As each $p_i$ is odd, the parity of the factor corresponding to $p_i$ is the same as the parity of $\alpha_i+1$, i.e., exactly one $\alpha_i$ is odd. That is, $n=p^\alpha N^2$, where $p$ does not divide $N$. It suffices to show that $N$ determines the prime power $p^\alpha$ and therefore for each $N^2\leq x$ there is only one odd perfect number of the above form. Now $$2=\frac{\sigma(n)}{n}=\left(1+\frac{1}{p}\cdots+\frac{1}{p^\alpha}\right)\frac{\sigma(N^2)}{N^2}$$ and so $$1+\frac{1}{p}+\cdots+\frac{1}{p^\alpha}=\frac{a}{b}$$ where $a/b$ is $2N^2/\sigma(N^2)$ in reduced form. As the denominator of the L.H.S. is $p^\alpha$, we must have $p^\alpha=b$.</p>