Jackson, Classical Electrodynamics - page 97: how to deal with the indexes of a series to obtain Legendre polynomials We want to solve Legendre's equation:
$$
   \frac{d}{dx}[(1-x^2)\frac{d}{dx}P(x)]+l(l+1)P(x)=0,\quad (1)
$$
Jackson writes $P(x)=x^\alpha\sum_{j=0}^\infty a_j x^j$, puts this in eqn. 1 and then comes up with:
$$
   \sum_{j=0}^\infty\{(\alpha+j)(\alpha+j-1)a_j x^{\alpha+j-2}-[(\alpha+j)(\alpha+j+1)-l(l+1)]a_jx^{\alpha+j}\}=0,\quad (2)
$$
and from this, he equals each term of the series to zero, and finds the recurrence relation below for the coefficients of the series:
$$
   \alpha_{j+2}=\frac{(\alpha+j)(\alpha+1+j)-l(l+1)}{(\alpha+j+1)(\alpha+j+2)}\cdot a_j,\quad (3)
$$
Now, I don't understand how he manages to obtain eqn. 2 from eqn. 1 and eqn. 3 from eqn. 2.
I rewrite eqn. 1 by doing the derivatives:
$$
   (1-x^2)\frac{d^2P}{dx^2}-2x\frac{dP}{dx}+l(l+1)P=0,\quad (1\text{bis})
$$
and then I evaluate the first and second derivatives of $P(x)$:
$$
   \frac{dP}{dx}=\frac{d}{dx}\sum_{j=0}a_jx^{\alpha+j}=\sum_{j=1}(j+\alpha)a_jx^{\alpha+j-1}=\sum_{j=0}(j+\alpha+1)a_{j+1}x^{\alpha+j}
$$
e
\begin{align}
   \frac{d^2P}{dx^2}&=\frac{d}{dx}\sum_{j=0}(j+\alpha+1)a_{j+1}x^{\alpha+j}\\
   &=\sum_{j=1}(\alpha+j+1)(\alpha+j)x^{\alpha+j-1}a_{j+1}\\
   &=\sum_{j=0}(\alpha+j+2)(\alpha+j+1)a_{j+2}x^{\alpha+j}
\end{align}
and put the whole in eqn. $1\text{bis}$:
\begin{align}
\sum_{j=0}[(j+\alpha+2)(j+\alpha+1)x^{j+\alpha}a_{j+2}&-(j+\alpha+2)(j+\alpha+1)x^{j+\alpha+2}a_{j+2}\\
   &-2(j+\alpha+1)x^{j+\alpha+1}a_{j+1}+l(l+1)a_jx^j]=0
\end{align}
what's next?
final edit: differentiation of the power series is wrong; see Marco81's answer for the correct development.
 A: Jackson starts from
$$\tag{1}
\frac{d}{dx}\Big[(1-x^2)\frac{dP}{dx}\Big]+l(l+1)P=0
$$
Substitution of  $P(x)=x^\alpha\sum_{j=0}^\infty a_j x^j$ in (1) will give:
$$\tag{2}
   \frac{dP}{dx}=\frac{d}{dx}\sum_{j=0}a_jx^{\alpha+j}=\sum_{j=0}(j+\alpha)a_jx^{\alpha+j-1}
$$
$$\tag{3}
(1-x^2)\frac{dP}{dx}=\sum_{j=0}(j+\alpha)a_jx^{\alpha+j-1}-\sum_{j=0}(j+\alpha)a_jx^{\alpha+j+1}
$$
$$\tag{4}
\frac{d}{dx}\Big[(1-x^2)\frac{dP}{dx}\Big]=\\
=\sum_{j=0}(j+\alpha)(j+\alpha-1)a_jx^{\alpha+j-2}-\sum_{j=0}(j+\alpha)(j+\alpha+1)a_jx^{\alpha+j}
$$
Putting all together in (1) you get:
$$\tag{5}
\sum_{j=0}(j+\alpha)(j+\alpha-1)a_jx^{\alpha+j-2}-\sum_{j=0}(j+\alpha)(j+\alpha+1)a_jx^{\alpha+j}+
\\+l(l+1)\sum_{j=0}a_jx^{\alpha+j}=0
$$
which gives Jackson's equation.
$$\tag{6}
\sum_{j=0}\{(j+\alpha)(j+\alpha-1)a_jx^{\alpha+j-2}-[(j+\alpha)(j+\alpha+1)-l(l+1)]a_jx^{\alpha+j}\}=0
$$
The two terms in the sum will have the same order when in the first "$j=j+2$" and this will give the relation you're looking for. Indeed:
$$\tag{7}
(j+\alpha+2)(j+\alpha+1)a_{j+2}x^{\alpha+j}-[(j+\alpha)(j+\alpha+1)-l(l+1)]a_jx^{\alpha+j}=0
$$
and 
$$\tag{8}
a_{j+2}=\frac{(j+\alpha)(j+\alpha+1)-l(l+1)}{(j+\alpha+2)(j+\alpha+1)}a_j
$$
