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 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.

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 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.

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.

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 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.