Exercise 1.4.7 (Series expansion of $1/(1-z)^{\alpha +1}$)

Question

Show that $\binom{-\alpha - 1}{k} = (-1)^k \binom{\alpha + k}{k}$ for $k\geq 0$. Deduce that if $|z| < 1$ then
$$\frac{1}{(1-z)^{\alpha + 1} }= \sum_{k=0}^\infty \binom{\alpha + k}{k} z^k.$$

Richard Ganaye · Accepted Answer

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\begin{proof} Let $\alpha \in \C$. By definition 1.6, for every $k\in \N$,
\begin{align*}
\binom{-\alpha - 1}{k} &= \frac{(-\alpha - 1)(-\alpha-2)\cdots(-\alpha - k)}{k!}\
&=(-1)^k \frac{(\alpha+k)\cdots (\alpha+2)(\alpha+1)}{k!}\\
&= (-1)^k \binom{\alpha + k}{k}.
\end{align*}
By 1.13, where $\alpha$ is replaced by $-\alpha-1$, and $z$ by $-z$, for every $z \in \C$ such that $|z|<1$,
\begin{align*}
\frac{1}{(1-z)^{\alpha + 1} }&= (1-z)^{-\alpha - 1}\
&=\sum_{k=0}^\infty \binom{-\alpha - 1}{k} (-z)^k\
&= \sum_{k=0}^\infty (-1)^k \binom{\alpha + k}{k} (-z)^k\
&= \sum_{k=0}^\infty  \binom{\alpha + k}{k} z^k.\
\end{align*}
\end{proof}

Exercise 1.4.7 (Series expansion of $1/(1-z)^{\alpha +1}$)

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Exercise 1.4.7 (Series expansion of $1/(1-z)^{\alpha +1}$)

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