Doubt in a derivation

Question

I am reading Hardy's book in divergent series and a step is making me question my intelligence. Page 178, equation (8.2.2), we have $\sum_{n=0}^{\infty}a_nx^{n+1}$ is a sequence converging to $f(x)$ for small enough $x$ and that $q>0$. He then defines a new variable $y$

$$ y=\frac{x}{1+qx}\qquad{}x=\frac{y}{1-qy} $$ and writes $$ \sum_{n=0}^{\infty}a_nx^{n+1}=\sum_{n=0}^{\infty}a_n\left(\frac{y}{1-qy}\right)^{n+1}=\sum_{n=0}^{\infty}a_n\sum_{m=n}^{\infty} \begin{pmatrix} m\\ n \end{pmatrix} q^{m-n}y^{m+1} $$

this last step confuses me. I know it has probbaly to do with the binomial theorem but the derivation eludes me. Aybody more enlightened than me?

Answer 1

We obtain with $x=\frac{y}{1-qy}$ \begin{align*} \sum_{n=0}^\infty x^{n+1}&=\sum_{n=0}^{\infty}a_n\left(\frac{y}{1-qy}\right)^{n+1}\\ &=\sum_{n=0}^{\infty}a_n\frac{y^{n+1}}{(1-qy)^{n+1}}\\ &=\sum_{n=0}^{\infty}a_ny^{n+1}\sum_{m=0}^{\infty}\binom{-(n+1)}{m}(-qy)^{m}\tag{1}\\ &=\sum_{n=0}^{\infty}a_ny^{n+1}\sum_{m=0}^{\infty}\binom{m+n}{n}(qy)^{m}\tag{2}\\ &=\sum_{n=0}^{\infty}a_ny^{n+1}\sum_{m=n}^{\infty}\binom{m}{n}(qy)^{m-n}\tag{3}\\ &=\sum_{n=0}^{\infty}a_n\sum_{m=n}^{\infty}\binom{m}{n}q^{m-n}y^{m+1}\tag{4}\\ \end{align*} and the claim follows.

Comment:

• In (1) we use the binomial series expansion.

• In (2) we use the binomial identity

\begin{align*} \binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^{q} \end{align*}

This identity holds since \begin{align*} \binom{-p}{q}&:=\frac{(-p)(-p-1)\cdots(-p-q+1)}{q!}\\ &=(-1)^q\frac{p(p+1)\cdots(p+q-1)}{q!}\\ &=(-1)^q\frac{(p+q-1)!}{(p-1)!q!}\\ &=(-1)^q\binom{p+q-1}{p-1} \end{align*}

• In (3) we shift the index $m$ to start from $n$ and substitute $m \rightarrow m-n$ in the sum accordingly.

• In (4) we collect the terms with $y$.