# Computing $\sum_{n=0}^{\infty} u_{n}, \frac{u_{n+1}}{u_n}=\frac{n+a}{n+b}$

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...?

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Hints: Assume first that $b\gt a+1$ and let $c=u_0\dfrac{b-1}{b-a-1}$.

• Show that $u_n=\displaystyle c\,(1-v_n)\prod_{k=0}^{n-1}v_k$ for every $n\geqslant0$, where $v_k=\dfrac{k+a}{k+b-1}$.
• Deduce that $\displaystyle\sum_{n=0}^Nu_n=c\,\left(1-\prod_{k=0}^{N}v_k\right)$ for every $N\geqslant0$.
• Deduce that $\displaystyle\sum_{n=1}^{+\infty}u_n=c$.

Extend this result to the fact that $\displaystyle\sum_{n=1}^{+\infty}u_n$ is infinite if $b\leqslant a+1$.

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