Let \(m\) be a fixed positive integer. Expand \((1-z)^{-m}\) in powers of \(z\). Also, if \[ (1-z)^{-m}=\sum_{n=0}^{\infty}a_nz^n, \] show that \[ a_n\sim\frac{1}{(m-1)!}n^{m-1} \qquad\text{as }n\to\infty. \]
Solution
1. For \(|z|<1\),\[ \frac1{1-z}=\sum_{k=0}^{\infty}z^k. \]
This is the geometric-series identity.
2. Differentiating the identity in 1 exactly \(m-1\) times gives\[ \frac{(m-1)!}{(1-z)^m} = \sum_{k=m-1}^{\infty}\frac{k!}{(k-m+1)!}z^{k-m+1}. \]
A power series may be differentiated term-by-term inside its radius of convergence. The \((m-1)\)st derivative of \((1-z)^{-1}\) is \((m-1)!(1-z)^{-m}\), while \[ \frac{d^{m-1}}{dz^{m-1}}z^k = \frac{k!}{(k-m+1)!}z^{k-m+1} \] for \(k\ge m-1\); lower-degree terms differentiate to zero.
3. Hence, for \(|z|<1\),\[ (1-z)^{-m} = \sum_{n=0}^{\infty}\binom{n+m-1}{m-1}z^n. \]
Set \(n=k-m+1\) in 2 and divide by \((m-1)!\). The coefficient becomes \[ \frac{(n+m-1)!}{n!(m-1)!} =\binom{n+m-1}{m-1}. \]
4. Therefore\[ a_n=\binom{n+m-1}{m-1} = \frac{(n+1)(n+2)\cdots(n+m-1)}{(m-1)!}. \]
3. The second is the factorial formula for the binomial coefficient; when \(m=1\), the numerator is the empty product and equals \(1\).
5. One has\[ \frac{a_n}{n^{m-1}/(m-1)!} = \prod_{j=1}^{m-1}\left(1+\frac jn\right) \longrightarrow1. \]
Divide the product expression in 4 by \(n^{m-1}/(m-1)!\). There are finitely many factors, and each tends to \(1\) as \(n\to\infty\).
6. Thus\[ a_n\sim\frac{n^{m-1}}{(m-1)!}. \]
This is exactly the ratio limit established in 5.