I would like to evaluate
$$ \sum_{n=0}^{\infty} u_{n}$$
where $u_{n}$ is defined by the following recurrence relation:
$$ \frac{u_{n+1}}{u_n}=\frac{n+a}{n+b}$$
$$ a,b>0$$ As $$ \frac{u_{n+1}}{u_n}=1-\frac{b-a}{n}+o(1/n) $$
a sufficient condition for the convergence of $\sum u_n$ is $b>a+1$
$$ u_{n}=\frac{(n-1+a)...(1+a)a}{(n-1+b)...(1+b)b}u_0=\frac{\Gamma(a+n)\Gamma(b)}{\Gamma(b+n)\Gamma(a)}u_0$$
So $$ \sum_{n=0}^{\infty} u_{n}=\sum_{n=0}^{\infty} \frac{\Gamma(a+n)\Gamma(b)}{\Gamma(b+n)\Gamma(a)}u_0$$
And...?