# Studying $\sum \frac{n!}{\prod_{k=1}^n (\alpha+k)}$, $\alpha>0$

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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Here's how I'd do it. We have

$$u_n=\frac{n!}{(1+\alpha)\cdots(n+\alpha)}=\frac{\Gamma(\alpha+1) \Gamma(n+1)}{\Gamma(n+\alpha+1)}.$$

By Gautschi's inequality we find that

$$(n+1)^{1-\alpha}<\frac{(n+1)u_n}{\Gamma(\alpha+1)} < (n+2)^{1-\alpha}$$

for $0<\alpha<1$. Using squeeze theorem, $p$-series test etc. we see that it diverges in this range. If we plug in $\alpha=1$ we get $u_n=1/(n+1)$, giving a harmonic series, so it doesn't converge here either.

For $1<\alpha<2$, set $s=\alpha-1$ so GI gives

$$(n+2)^{1-s}<\frac{(n+1)(n+2)u_n}{\Gamma(\alpha+1)}<(n+3)^{1-s}.$$

Comparing with $\zeta(1+s)$, we see that the series converges here. Finally, for $\alpha>2$, observe that each term in the series $u_n$ decreases as $\alpha$ increases, so convergence in $(1,2)$ entails convergence on the entire interval $(1,\infty)$, with the appropriate bookkeeping.

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@Eric: we should get $\displaystyle \frac{(n+\bullet)^{1-s}}{(n+1)(n+2)}\sim \frac{1}{n^{1+s}},$ hence we compare to $\zeta(1+s)$. –  anon Feb 22 '12 at 10:06
Ahhh ok! Thanks. –  Eric Naslund Feb 22 '12 at 12:24