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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...?

2 Answers 2