The series $\sum_n\Gamma (n-1/3)/(n-1)!$ diverges

Question

I would like to prove that the series: $$\sum_{n=1}^{\infty}\frac{\Gamma (n-1/3)}{(n-1)!}$$ diverges. The problem is that I don't know how to begin.

Intuitively I get the result, because observing the terms of the series as they're summed up the sum gets bigger and bigger and it blows up.

Any ideas would be useful.

Answer 1

Hint: show that

$$\frac{c}{n^{1/3}}\le\frac{\Gamma (n-1/3)}{(n-1)!},$$

for some positive constant $c>1$ and use the comparison test. $c=2$ works.

Answer 2

It seems that $$\sum_{n=1}^{m}\frac{\Gamma (n-1/3)}{(n-1)!}=\frac{3 \Gamma \left(m+\frac{2}{3}\right)}{2 \Gamma (m)}$$ May be, you could use induction to prove it and go to the limit.

If you approximate the rhs using Stirling approximation, you end with $$\frac{3 \Gamma \left(m+\frac{2}{3}\right)}{2 \Gamma (m)} \simeq \frac{3 (m-1)^{\frac{1}{2}-m} \left(m-\frac{1}{3}\right)^{m+\frac{1}{6}}}{2 e^{2/3}} > \frac{3 (m-1)^{2/3}}{2 e^{2/3}}$$

Answer 3

Let

$$u_n=\frac{\Gamma(n-1/3)}{(n-1)!}$$ then by the relation $$\Gamma(x+1)=x\Gamma(x)$$ we have $$\frac{u_{n+1}}{u_n}=\frac{n-1/3}{n}=1-\frac{\frac13}{n}$$ hence by the Raabe-Duhamel's rule the series is divergent.

Answer 4

For every $a\gt-1$, $b\gt0$, $b\ne a+1$, one has $$\sum_{k=1}^n\frac{\Gamma (k+a)}{\Gamma(k+b)}=\frac1{a+1-b}\left(\frac{\Gamma(n+a+1)}{\Gamma(n+b)}-\frac{\Gamma(a+1)}{\Gamma(b)}\right).$$ Taking the limit $b\to0$ yields, for every $a\gt-1$, $$\sum_{k=1}^n\frac{\Gamma (k+a)}{(k-1)!}=\frac1{a+1}\frac{\Gamma(n+a+1)}{(n-1)!}.$$ The formula in @ClaudeLeibovici's post is the case $a=-1/3$.