Frobenius expansion of the Legendre equation at an endpoint ($a_{k+1}=\frac{(k-n)(k+n+1)}{2(k+1{)}^2}{a}_k$)

For the Legendre differential equation $(1-x^2)y^{\prime \prime }-2x{y}^{\prime }+n(n+1)y=0$, put $t=1+x$. The equation becomes $t(2-t)y_{tt}+2(1-t)y_t+n(n+1)y=0$. Its indicial equation is $2r^2=0$. The exponent-zero power series $\sum a_k{t}^k$ obeys the displayed linear recurrence relation. For a nonnegative integer $n$, it terminates at degree $n$, giving a Legendre polynomial with $a_0\ne 0$. Every analytic solution at the endpoint is proportional to this one, since the recurrence determines all coefficients from $a_0$.

The second independent solution has the local form $P_n(x)\log (1+x)+B(x)$, where $B$ is analytic near $-1$. Indeed, reduction of order gives $P_n(x)\int dx/[(1-x^2)P_n(x{)}^2]$, whose integrand is $1/[2P_n(-1{)}^2(1+x)]$ plus an analytic function. Rescale the second solution to obtain the unit coefficient of $P_n\log (1+x)$. Reflection gives the corresponding expansion at $x=1$. This also proves the logarithmic divergence of the independent Legendre function of the second kind at each endpoint.