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##\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