How to show that the Legendre Series solution $y_{even}$ and $y_{odd}$, diverges as $x = \pm1 $.
$y_{even} = \sum_{j=0,2,\ldots}^\infty a_jx^j$, where $a_{j+2}=\frac{j(j+1)-n(n+1)}{(j+1)(j+2)}a_j$.
$y_{odd}=\sum_{j=1,3,\ldots}^\infty a_jx^j$, where $a_{j+2}=\frac{(j+1)(j+2)-n(n+1)}{(j+2)(j+3)}a_j$.
My Work:
I have shown that for both the $y_{even}$, and the $y_{odd}$, solutions, that $\lim |\frac{a_{j+1}}{a_j}|$, as $j \rightarrow \infty$ is $1$.
I cannot find the next part? How to proceed?
P.S. This problem is from Arfken. Problem #8.3.1 (7th Ed.)
Update: I have been thinking about this for a while, and I have come up with this. I am using the Gauss test.
I am taking the $y_{even}$ solution, and defining $\alpha = \left|\frac{a_{2j}}{a_{2j+2}}\right| = \frac{(2j+1)(2j+2)}{(2j-n)(2j+n-1)}$.
Now after some manipulation, I get my $\alpha$ to be,
$$\alpha=1+\frac1j+\frac{n(n+1)(j+1)}{(4j^2+2j-n(n+1))j}$$
Noe according to the Gauss test, one has to check that my third term, viz., $\frac{n(n+1)(j+1)}{(4j^2+2j-n(n+1))j}$, which, I should be able to manipulate to $\frac{C_j}{j^r}$; for $\lim_{j\to\infty}C_j$, should be finite, then only I can say that, the series does diverge for $x\rightarrow\pm1$.
Okay, thought for some more time, and I have a solution to this problem. Now I can manipulate the third term in $\alpha$, which would lead me to,
$$\frac{n(n+1)(1+\frac1j)}{j^2 \left(4+\frac2j-\frac{n(n+1)}{j^2} \right) }$$
which I can identify straight as my $\frac{c_j}{j^r}$ term in my Gauss test. So here $r=2$, and $c_j=\frac{n(n+1)(1+\frac1j)}{\left(4+\frac2j-\frac{n(n+1)}{j^2} \right) }$. And when I do take the $\lim_{j\to\infty}c_j=\frac{n(n+1)}4$, so it does remain bounded.
So now I can draw a conclusion about the convergence/divergence at $x\rightarrow\pm1$ for the Legendre ODE. Now my $h=1$, and $r(\equiv2)>1$, and also $\lim_{j\to\infty}=\frac{n(n+1)}4$, so I can say with confirmation, that at $x=\pm1$, the series diverges!
This completes the proof, I guess.
See About the Legendre differential equation where the recurrence relations above are derived from entering $y=\sum_{j=0}^\infty a_jx^j$ into the Legendre differential equation.
Solutions to the equation are $y=P_n(x)$ which is a polynomial of degree $n$, and $y=Q_n(x)$, which is constructed of polynomials and logarithms.
$P_n(1)=1$ and $P_n(-1)=\pm 1$ so in these cases the series does not diverge.
But indeed the $Q_n(x)$ series diverges for $x=\pm 1$.
For example: $Q_0(x)=\frac{1}{2} ln(\frac{1+x}{1-x})$