Convergence of $ u_{n}=\sqrt [n]{\frac{(a+1)(a+2)...(a+n)}{n!}} $ I would like study the convergence of the following sequence:
$$ u_{n}=\sqrt [n]{\frac{(a+1)(a+2)...(a+n)}{n!}} $$
where $a>0$
We have:
$$ \ln(u_{n})=\frac{1}{n}\sum_{k=1}^n \ln(1+\frac{a}{k})$$
And : $$ \ln(u_{n+1})-\ln(u_{n})=\frac{1}{n+1}\sum_{k=1}^{n+1} \ln(1+\frac{a}{k})-\frac{1}{n}\sum_{k=1}^n \ln(1+\frac{a}{k})$$ 
So I have to study the convergence of $$ \sum \ln(u_{n+1})-\ln(u_{n})$$ 
Using integrals:  $$ \sum_{k=1}^n \ln(1+\frac{a}{k}) \sim a\ln(n)$$
Thus: $$ \ln(u_{n+1})-\ln(u_{n})= \frac{1}{n}a\ln(n)-\frac{1}{n}a\ln(n)+o(\frac{\ln(n)}{n})=o(\frac{\ln(n)}{n}) $$
which is not enough to determine the convergence or divergence of $ \sum \ln(u_{n+1})-\ln(u_{n})$
So how can I find a more precise approximation of $ \ln(u_{n+1})-\ln(u_{n})$?
Or is there a simple equivalent of $$ \prod_{k=1}^n {(a+k)}$$ ?
 A: Just a note on evaluating the limit of your sequence:
You can compute the limit by using the fact that if $\lim\limits_{n\rightarrow\infty}{a_{n+1}\over a_n} $ exists, then so does $\lim\limits_{n\rightarrow\infty}{\root n\of {a_n}}$ and the two limits are equal.
Set $a_n={(a+1)(a+2)\cdots(a+n)\over n!}$. Then a simple computation shows that $\lim\limits_{n\rightarrow\infty}{a_{n+1}\over a_n}=1 $.  So, 
$\lim\limits_{n\rightarrow\infty}{\root n\of{ a_n}}
=\lim\limits_{n\rightarrow\infty}{\root n\of {(a+1)(a+2)\cdots(a+n)\over n!}}
=1$. 
A: Since $ \log(1+x) \leq x$ and $\displaystyle \sum_{k=1}^n \frac{1}{k} \sim \log n, $ $$ \ln(u_{n})=\frac{1}{n}\sum_{k=1}^n \log \left(1+\frac{a}{k}\right) \leq \frac{a}{n}\sum_{k=1}^n\frac{1}{k} \to 0$$
thus $u_n \to 1.$
A: Another proof, which uses the $\text{GM} \le \text{AM}$ inequality.
We have
$$1 \le \sqrt[n]{\frac{(a+1)}{1}\cdot \frac{(a+2)}{2} \cdots \frac{(a+n)}{n}} \le \frac{1}{n}\sum_{k=1}^{n} (1 + \frac{a}{k}) = 1 + \frac{1}{n} \sum_{k=1}^{n} \frac{a}{k}$$
Since $\frac{a}{n} \to 0$, we have that $\frac{1}{n} \sum_{k=1}^{n} \frac{a}{k} \to 0$ and so your sequence converges to $1$.
