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I would like to find for which values of $ \alpha>0 $ the series $ \sum {u_{n}} $ is a convergent series, where $$ u_{n}= \frac{n!}{\prod_{k=1}^n (\alpha+k)}$$

Here is what I have done:

$$ \frac{u_{n+1}}{u_{n}}=1-\frac{\alpha}{\alpha+n+1}$$

We can introduce the Riemann series $ \sum v_{n}$ where $$ v_{n}=\frac{1}{(n+1+\alpha)^{\beta}}$$

$$ \frac{v_{n+1}}{v_{n}}=1-\frac{\beta}{\alpha+n+1}+o(1/n)$$

If $ \alpha>1 $ and $1<\beta<\alpha$

$$ \frac{u_{n+1}}{u_{n}}-\frac{v_{n+1}}{v_{n}}= \frac{\beta-\alpha}{n+1+\alpha}+o(1/n) $$

$$ \exists N\in \mathbb{N}/ \forall n\geq N: \frac{u_{n+1}}{u_{n}} \leq \frac{v_{n+1}}{v_{n}} $$

As $ \sum v_{n} $ is a convergent series, $ \sum u_{n} $ is also convergent.

If $ \alpha<1 $ and $ \alpha<\beta<1$

$$ \exists N'\in \mathbb{N}/ \forall n\geq N': \frac{u_{n+1}}{u_{n}} \geq \frac{v_{n+1}}{v_{n}} $$

As $\sum v_{n} $ is a divergent series, $ \sum u_{n} $ is also divergent.

$ \alpha=1 $: $ u_{n}=\frac{1}{n+1}$, $ \sum u_{n} $ is a divergent series.

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