# Limit of a summation wth Gamma function

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 $\alpha>3$.

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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).
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