Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $$ \sum_{n=0}^{\infty} u_{n}=\sum_{n=0}^{\infty} \frac{\Gamma(a+n)\Gamma(b)}{\Gamma(b+n)\Gamma(a)}u_0$$


share|cite|improve this question

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 $b\leqslant a+1$.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.