Q:

Prove $\sum\limits_l\frac{R^{2l+1}r^{-l-1}P_l(\cos\theta)}{a^{l+1}}=\frac{R}{a\sqrt{(R^2/a)^2-2r\cos\theta (R^2/a)+r^2}}$ from LHS $\to$ RHS

This is related to this previous question of mine where (with lots of help) I show that $$\sum_{l=0}^\infty \frac{R^l}{a^{l+1}}P_l(\cos\theta)=\frac{1}{\sqrt{a^2-2aR\cos\theta+R^2}}\tag{1}$$ by using the Legendre generating function.
The generating function for Legendre Polynomials is:
$$\Phi(x,h)=(1-2xh+h^2)^{-1/2} \quad\text{for}\quad |h|\lt 1\tag{2}$$
or
$$\Phi(x,h)=\sum_{l=0}^\infty h^l P_l(x)\quad\text{for}\quad |h|\lt 1\tag{3}$$
I need to show that $$\sum_{l=0}^\infty\frac{R^{2l+1}r^{-l-1}P_l(\cos\theta)}{a^{l+1}}=\frac{R}{a\sqrt{(R^2/a)^2-2r(R^2/a)\cos\theta+r^2}}\tag{4}$$ I note the striking similarity between the RHS's of $(1)$ and $(4)$ within the square root of the denominator.

It is not necessary to read the following if you understand how to prove that the sum can be written as above; But here is some context anyway:

 

So here is my attempt:
From the RHS of equation $(4)$:
$$\displaystyle\begin{align}\frac{R}{a\sqrt{r^2-2r(R^2/a)\cos\theta+(R^2/a)^2}}&=\frac{R}{ar\sqrt{1-2(R^2/ar)\cos\theta+(R^2/ar)^2}}\end{align}$$
Changing variables to match equation $(2)$: Let $h=R^2/ar$ and $x=\cos\theta$
Then 
$$\displaystyle\begin{align}\frac{R}{ar\sqrt{1-2(R^2/ar)\cos\theta+(R^2/ar)^2}}&=\frac{R}{ar\sqrt{1-2hx+h^2}}\\&=\frac{R}{ar}(1-2xh+h^2)^{-1/2}\\&=\frac{R}{ar}\Phi(x,h)\\&=\frac{R}{ar}\sum_{l=0}^\infty h^l P_l(x)\\&=\frac{R}{ar}\sum_{l=0}^\infty \frac{R^{2l} P_l(\cos\theta)}{(ar)^l}\\&=\sum_{l=0}^\infty \frac{R^{2l+1} P_l(\cos\theta)}{a^{l+1}r^{l+1}}\\&=\sum_{l=0}^\infty \frac{R^{2l+1}r^{-l-1} P_l(\cos\theta)}{a^{l+1}}\quad\fbox{}\end{align}$$

You may be wondering what I'm asking at this point; so here is the question:
In the last line of the extract the textbook says "by summing the series"; and this seems to suggest that one must arrive at the RHS from the LHS (instead of going from RHS to LHS as I did). So is my interpretation correct (LHS $\to$ RHS) and if so can it be done in this order?
Many thanks.

A:

Let's start with an interpretation of the paragraph:
\begin{align*}
\frac{q}{\sqrt{R^2-2aR\cos \theta+a^2}}
&=q\sum_{l}\frac{R^lP_l(\cos\theta)}{a^{l+1}}\qquad\qquad\quad\ \qquad\qquad (8.18)\\
\\
\\
V=\frac{q}{\sqrt{r^2-2ar\cos\theta+a^2}}&
-q\sum_{l}\frac{R^{2l+1}r^{-l-1}P_l(\cos \theta)}{a^{l+1}}
\qquad\qquad\qquad (8.20)
\end{align*}

Since the second term in (8.20) is of the same general form as (8.18), we can simplify (8.20) by summing the series to get (8.21).
  \begin{align*}
V=\frac{q}{\sqrt{r^2-2ar\cos\theta+a^2}}&
-\frac{\left(\frac{R}{a}\right)q}
{\sqrt{r^2+\left(\frac{R^2}{a}\right)^2-2r\left(\frac{R^2}{a}\cos\theta\right)}}\qquad\qquad (8.21)
\end{align*}

The intention of the stated paragraph is to take a look at the series in (8.20), observe the similar form with the RHS of (8.18) and use a closed form corresponding to the LHS of (8.18).
We see the series in (8.20) is $\frac{R}{r}$ times the series in (8.18) when replacing $R$ with $\frac{R^2}{r}$. So we can take the closed form of (8.18) substitute $R$ with $\frac{R^2}{r}$, multiply it with $\frac{R}{r}$ and we are done.

We obtain
  \begin{align*}
\sum_{l}\frac{R^{2l+1}r^{-l-1}P_l(\cos \theta)}{a^{l+1}}&=
\frac{R}{r}\frac{q}{\sqrt{\left(\frac{R^2}{r}\right)^2-2a\frac{R^2}{r}\cos \theta+a^2}}\\
&=\frac{\left(\frac{R}{a}\right)q}
{\sqrt{\left(\frac{R^2}{a}\right)^2-2r\left(\frac{R^2}{a}\cos\theta\right)+r^2}}
\end{align*}
  according to (8.21).

Note: Since there is always more than one way to do it, your interpretation is correct as well. ... and it can be done in the order you did. The essence of the stated para is to earn from the relationship in (8.18) to easily obtain (8.20). So, from my point of view, both variants are correct and quite ok.

