1

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.

Chon
  • 6,002

1 Answers1

2

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.

Community
  • 1
anon
  • 151,657