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).