Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am interested in the asymptotic order of the partition function $p(n)$.

The paper Asymptotic Formulae in Combinatory Analysis proves there are constants $A$,$B$ such that $e^{A\sqrt{n}} < p(n) < e^{B\sqrt{n}}$ by elementary means. Here is the argument for one side of the inequality, it is just the inductive step which I have not understood:

I assume that $p_r(n)$ is defined to be $0$ for negative $n$, then the inequalty (2.22) does not hold for these $n$. In which case the sum $\{n^{s-1} + (n-s-1)^{s-1} + (n-2s-2)^{s-1} + \ldots\}$ must be finite, ending before $n-ks-k$ becomes negative. On the other hand the use of telescoping down to $n^s$ suggests that the sum is infinite, otherwise we would end up with $n^s - (n-ks-k)^{s-1}$ and are unable to throw away the $k$ term.

Thank you to anyone who will help me understand this argument.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Write $n = Q(s-1)+R$ with $0 \le R < s-1$. The final term in the first line of the induction step (i.e. the three-line display equation) is $R^{s-1}$. As you have noted, as written the author seems to imply that the final term in the next line of the inequality should be $(R^s - (R-(s-1))^s)/s(s+1)$. But in fact $R^s/s(s+1)$ will do, because $R^{s-1} \ge R^s/s(s+1)$ (recall that $R < s-1$), and then your sum telescopes just fine.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.