I know that any second order linear ode $$w''+p(z)w'+q(z)w=0$$ is of Fuchs type (ie, coefficients meromorphic with singular points at $z_0,...,z_n,\infty$ all regular) if $$p(z)=\sum\frac{p_j}{z-z_j}, q(z)=\sum \frac{q_j}{(z-z_j)^2}+\frac{r_j}{z-z_j},\quad \sum r_j=0$$ where the sums are over the singular points $z_0,...,z_n$.
I wish to show that there is no singularity at $\infty$ if $$2-\sum p_j=\sum (q_j+r_jz_j)=\sum z_j(2q_j+r_jz_j)=0$$
To do this, I'm guessing that I have to make the substitution $u=1/z$ to obtain $$w''(u)+(2/u-p(1/u)/u^2)w'(u)+1/u^4q(1/u)w(u)=0$$ but now I'm getting stuck. Anyone have any insights?
It is always helpful to not reuse the same variable names for a different object. Define $\tilde w(u)=w(1/u)$, then $$ \tilde w'(u)=-\frac1{u^2}w'(1/u),~~~ \tilde w''(u)=\frac1{u^4}w''(1/u)+\frac{2}{u^3}w'(1/u), $$ so that the transformed equation is $$ 0=u^4\tilde w''(u)-2u^3\tilde w'(u)-p(1/u)u^2\tilde w'(u)+q(1/u)w(u), \\~\\ 0=\tilde w''(u)-\left(\frac2u+\frac{p(1/u)}{u^2}\right)\tilde w'(u)+\frac{q(1/u)}{u^4}w(u) $$ Now the solution $w$ extends to infinity if the coefficients in the last form of the equation are at least continuous in and around $u=0$, in our context this automatically implies holomorphic around $u=0$.
For $$ \frac1u\left(2+\sum\frac{p_j}{1-z_ju}\right) $$ to be continuous in $u=0$ the second factor has to evaluate to zero at $u=0$, giving the first claimed condition.
The second coefficient is $$ \frac1{u^2}\sum \frac{q_j}{(1-z_ju)^2}+\frac1{u^3}\sum \frac{r_j}{1-z_ju} \\ = \frac1{u^2}\sum q_j(1+2(z_ju)+3(z_ju)^2+...)+\frac1{u^3}r_j(1+z_ju+(z_ju)^2+(z_ju)^3...) $$ using geometric or binomial series. Now collect terms of equal degree and demand that the coefficients of negative powers vanish, $$ 0=\sum r_j=\sum (q_j+r_jz_j)=\sum z_j(2q_j+r_jz_j). $$ Was there a transmission error in the middle term?