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...?