 Frobenius recurrence for the Legendre equation

Substitute $y=x^{\sigma }{\sum }_{n\ge 0}{a}_n{x}^n$, $a_0\ne 0$, into the Legendre differential equation. The lowest powers give $\sigma (\sigma -1)=0$ and $\sigma (\sigma +1)a_1=0$; subsequent powers give the recurrence in the title. Thus exponent zero has independent even and odd coefficients, while exponent one requires $a_1=0$ and represents the odd solution. Zero is an ordinary point of the equation, so the exponent-one choice simply selects a solution that vanishes there.

The singular points at $\pm 1$ bound the generic radius of convergence about zero by one. Every nonterminating parity recurrence has successive-coefficient ratio tending to one in the variable $x^2$, giving radius exactly one. If $\lambda =N$ is a nonnegative integer, choosing the parity of $N$ makes the recurrence terminate at degree $N$, producing a Legendre polynomial with infinite radius. These are precisely the nonnegative parameters with a nonzero entire solution: even and odd parts of an entire solution would both have to be entire, so any nonzero nonterminating part is excluded.