Is power series expansion not allowed for Gauss' Convergence Test?

Question

Given the following Legendre series, I wanted to test its convergence at $x^2=1$ when $l$ is not an integer \begin{align} S = \sum_{j=0}^{\infty} u_j && u_{j+2} = \dfrac{(j+1)(j+2)-l(l+1)}{(j+2)(j+3)} x^2 u_{j} \end{align} I tried using Gauss' test for the following Legendre Series \begin{align} \lim\limits_{j \rightarrow \infty} \left| \dfrac{u_{j}}{u_{j+2}} \right|_{x^2 = 1} = \lim\limits_{j \rightarrow \infty} \dfrac{(j+2)(j+3)}{ (j+1)(j+2)-l(l+1)} = \\ \lim\limits_{j \rightarrow \infty} \dfrac{ j^2 + 5j+6 }{ j^2+3j+2-l(l+1) } = \lim\limits_{j \rightarrow \infty} \dfrac{ [j^2+3j+2-l(l+1)] +[2j+4+l(l+1)] }{ j^2+3j+2-l(l+1) } \\= \lim\limits_{j \rightarrow \infty} 1+ \dfrac{ 2j+4+l(l+1) }{ j^2+3j+2-l(l+1) } = 1 + \dfrac{2}{j} + \mathcal{O}(j^{-2}) = 1 + \dfrac{h}{j} + \mathcal{O}(j^{-2}) \end{align} I identify the value of Gauss' coefficient $h$ from the power series expansion of the of the ratio of the two adjacent series coefficients as illustrated above. I keep getting a factor of 2 for Gauss' coefficient $h$ which states that the series convergence, however my friends tell me that the series diverges. Is it possible that power series expansion cannot be used to evaluate Gauss' coefficient?

Answer 1

Your mistake is to apply a recipe valid when the ratios $$\left|\frac{u_j}{u_{j+1}}\right|$$ are considered, to a setting where you estimate the ratios $$\left|\frac{u_j}{u_{j+2}}\right|.$$ Renaming $u_{2k}=v_k$, your computations yield $$\left|\frac{v_k}{v_{k+1}}\right|=1+\frac2{2k}+O\left(\frac1{k^2}\right)=1+\frac{\color{red}{1}}{k}+O\left(\frac1{k^2}\right),$$ that is, the case $c=\color{red}{1}$ of the recipe you are trying to apply. (Hence you need to get one more term in the expansion to be able to conclude but at least now, the prediction of your friends is a possible outcome.)