Raabe's test says Legendre functions always converge?

Tags:
1. May 13, 2015

Happiness

The Legendre functions are the solutions to the Legendre differential equation. They are given as a power series by the recursive formula (link [1] given below):

\begin{align}y(x)=\sum_{n=0}^\infty a_n x^n\end{align}

\begin{align}a_{n+2}=-\frac{(l+n+1)(l-n)}{(n+1)(n+2)}a_n\end{align}

If $l$ is a non-negative integer, The Legendre functions reduce to the Legendre polynomials. Otherwise, they are divergent (for $x\in[-1,1]$).

However, I did Raabe's test and found that they are convergent for all values of $l$, a wrong result.

My workings:

\begin{align}y(1)&=\sum_{n=0}^\infty a_n\,1^n\\ &=\sum_{n=0,2,4...}^\infty a_n+\sum_{n=1,3,5...}^\infty a_n\end{align}

Let $\,y_1=\sum_{n=0,2,4...}^\infty a_n\,\,\,\,\,\,\,\,$ and $\,\,\,\,\,\,\,\,y_2=\sum_{n=1,3,5...}^\infty a_n$

Applying Raabe's test to $\,y_1$ (note that the $\,a_{n+2}$ term in (2) becomes the $\,a_{n+1}$ term in $y_1$),

\begin{align}R&=\displaystyle\lim_{n\rightarrow +\infty}n\big(\big|{\frac{a_n}{a_{n+1}}}\big|-1\big)\\ &=\displaystyle\lim_{n\rightarrow +\infty}n\big[\frac{(n+1)(n+2)}{(l+n+1)(n-l)}-1\big]\\ &=\frac{2n^2+\big(l(l+1)+2\big)n}{n^2+n-l(l+1)}\\ &=2\end{align}

$R>1$. Thus, $\,y_1$ is absolutely convergent.

Since both $\,y_1$ and $\,y_2$ follows the same recursive formula (2), $\,y_2$ is also absolutely convergent. So $\,y(1)$ is absolutely convergent. Since $\,y(1)$ is absolutely convergent, $\,y(-1)$ is convergent.

But $\,y(1)$ or $\,y(-1)$ must diverge. Where's my mistake?

Background information:

http://mathworld.wolfram.com/LegendreDifferentialEquation.html

http://en.wikipedia.org/wiki/Ratio_test

2. May 18, 2015