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

Can anyone prove this (I'm very confident that it is correct) or have any idea how this can be handled:

$$ \lim_{n \rightarrow \infty} \frac{1}{n-1}\sum_{i=1}^{n-1} \frac{1}{(\alpha-1)(n-i) -1} \frac{n!}{(n-i-1)!} \frac{\Gamma(n-i+1-2/(\alpha-1))}{\Gamma(n+1-2/(\alpha-1))} = \frac{1}{\alpha-3}, $$

with $\alpha>3$.

share|cite|improve this question
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – Julian Kuelshammer Oct 15 '12 at 21:05
up vote 3 down vote accepted

Replacing the prefactor $1/(n-1)$ by $1/n$ does not change the limit. Using the change of variable $k=n-i$ and the shorthand $c=1/(\alpha-1)$, one gets that the $n$th quantity is $$ S_n=\frac{c\Gamma(n)}{\Gamma(n+1-2c)}\sum_{k=1}^{n-1}\frac{\Gamma(k+1-2c)}{(k-c)\Gamma(k)}. $$ The prefactor is equivalent to $cn^{2c-1}$ and, when $k\to\infty$, the $k$th term in the sum is equivalent to $k^{-2c}$, hence there are three regimes:

  • If $c\gt\frac12$, then the sum over $k$ converges to a finite limit and the prefactor goes to $+\infty$, hence $S_n\to+\infty$. This occurs when $1\lt\alpha\lt3$.
  • If $c=\frac12$, then the sum over $k$ goes to $+\infty$ and the prefactor is constant equal to $c=\frac12$, hence $S_n\to+\infty$. This occurs when $\alpha=3$.
  • If $c\lt\frac12$, then the prefactor goes to zero hence the exact values of the first terms in the sum do not matter. What matters is that the $k$th term is equivalent to $k^{-2c}$ hence a comparison with a Riemann integral shows that the whole sum is equivalent to $\int\limits_0^nt^{-2c}\mathrm dt=\frac{n^{1-2c}}{1-2c}$. Multiplying this by the prefactor yields $S_n\to\frac{c}{1-2c}=\frac1{\alpha-3}$. This occurs when $\alpha\gt3$.

Finally, the result holds for every $\alpha\gt3$ (and not for every $\alpha\gt2$, as previously, erroneously, claimed in the question).

share|cite|improve this answer
Awesome. Thanks a lot. The property that $\lim_{n\rightarrow \infty} \frac{\Gamma (n+\alpha)}{\Gamma (n) n^\alpha}=1$ is essential. – RichardKwo Oct 5 '12 at 1:52

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.