Limit of $\sqrt[n]{(x+1)...(x+n)} - x$ as $x \to +\infty$ Let $n \in \mathbb{N}^{\ast}$. I want to determine the following limit :
$$ \lim \limits_{x \to +\infty} \sqrt[n]{(x+1)\ldots(x+n)} - x.$$
Let $x = \frac{1}{t}$ with $t \to 0$. It is equivalent to the following limit :
$$ \lim \limits_{t \to 0} \sqrt[n]{\displaystyle \frac{(t+1)\ldots(t+n)}{t^{n}}} - \frac{1}{t}. $$
$$
\begin{align*}
\sqrt[n]{\displaystyle \frac{(t+1)\ldots(t+n)}{t^{n}}} - \frac{1}{t} &= {} \frac{1}{t} \sqrt[n]{(t+1)\ldots(t+n)} - \frac{1}{t} \\[2mm]
 &= \frac{1}{t} \sqrt[n]{ n! + u(t) } - \frac{1}{t} \\[2mm]
\end{align*}
$$
where $u(0)=0$ and $\displaystyle \lim \limits_{t \to 0} u(t) = 0$. Since $\displaystyle \sqrt[n]{1+t} = 1 + \frac{t}{n} + o(t)$ as $t \to 0$, $$\displaystyle \sqrt[n]{n! + u(t)} = \sqrt[n]{n!} \; \sqrt[n]{\displaystyle 1 + \frac{u(t)}{n!} } = \sqrt[n]{n!} \bigg( 1 + \frac{u'(0) t}{n(n!)} + o(t) \bigg).$$
I do not see how to go on from there !

$\bf{My\; Solution::}$ Given $$\displaystyle \lim_{x\rightarrow \infty}\sqrt[n]{(x+1)(x+2)(x+3).....(x+n)}-x$$
Using $\bf{A.M\geq G.M},$ Here $$(x+1),(x+2),(x+3),............(x+n)>0\;,$$ when $x\rightarrow \infty$
So $$\displaystyle \frac{(x+1)+(x+2)+..............+(x+n)}{n}\geq \sqrt[n]{(x+1)(x+2)............(x+n)}$$
and equality hold when $$(x+1)=(x+2)=.............=(x+n)\;,$$ bcz here $x\rightarrow \infty$
So all are equal, 
So $$\displaystyle \frac{nx+\frac{n(n+1)}{2}}{n} = \sqrt[n]{(x+1)(x+2)...........(x+n)}$$
So $$\displaystyle \lim_{n\rightarrow \infty}\frac{nx+\frac{n(n+1)}{2}}{n}-x = \lim_{n\rightarrow \infty}\sqrt[n]{(x+1)(x+2)...........(x+n)}-x$$ 
So $$\displaystyle \lim_{n\rightarrow \infty}\sqrt[n]{(x+1)(x+2)...........(x+n)}-x = \frac{n+1}{2}$$
