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 !

We have
$$\sqrt[n]{a} - \sqrt[n]{b} = \dfrac{a-b}{\sum_{k=0}^{n-1} a^{k/n}b^{(n-1-k)/n}}$$
In our case, $a = (x+1)(x+2)\cdots(x+n)$ and $b=x^n$. We have $a-b = \frac{n(n+1)}2x^{n-1} + O(x^{n-2})$. The denominator is $nx^{n-1} + O(x^{n-2})$. Hence, we have the expression as
$$\frac{n(n+1)/2 + O(1/x)}{n + O(1/x)} \to \frac{(n+1)}2 \text{ as }x \to \infty$$
