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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...?

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2 Answers 2

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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You may find the same question in the book T.J.I'A. Bromwich, "An introduction to the theory of infinite series" (1947), second edition, p.48. The proof is different from did's one. You may see also the problems of Amer.Math.Monthly num. 11260, 11409, 11473

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