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edited Sep 1 2010 at 10:42

complete estimates of the error for a well-known asymptotic expression in of partition p(n,m)

Let $p(n,m)$ be the number of partitions of an integer $n$ into integers $\le m$, we have a well-known asymptotic expression:

For a fixed $m$ and $n\to\infty$, $$p(n,m)=\frac{n^{m-1}}{m!(m-1)!} (1+O(1/n)) $$

My question is: why the error $O(1/n)$ is indepdent independent of $m$? Or how can it be extended for $m$ growing slowly with $n$? Please help me to find the answer or the references. Thanks.

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edited Sep 1 2010 at 10:13

Robin Chapman
14.5k●2●29●50

Let p(n,m) $p(n,m)$ be the number of partitions of an integer n $n$ into integers <= m$\le m$, we have a well-known asymptotic expression:

For a fixed m $m$ and n-> infinity, p(n,m)=n^(m-1)/(m!(m-1)!) $n\to\infty$, $$p(n,m)=\frac{n^{m-1}}{m!(m-1)!} (1+O(1/n)) $$

My question is: why the error O(1/n) $O(1/n)$ is indepdent of m? $m$? Or how can it be extended for m $m$ growing slowly with n? $n$? Please help me to find the answer or the references. Thanks.

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asked Sep 1 2010 at 9:43

QHLIU
59●6

complete estimates of the error for a well-known asymptotic expression in partition p(n,m)

Let p(n,m) be the number of partitions of an integer n into integers <= m, we have a well-known asymptotic expression:

For a fixed m and n-> infinity, p(n,m)=n^(m-1)/(m!(m-1)!) (1+O(1/n))

My question is: why the error O(1/n) is indepdent of m? Or how can it be extended for m growing slowly with n? Please help me to find the answer or the references. Thanks.