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 :

$$ \lim \limits_{t \to 0} \sqrt[n]{\displaystyle \frac{(t+1)\ldots(t+n)}{t^{n}}} - \frac{1}{t}. $$

$$
\begin{align*}
\sqrt[n]{\displaystyle \frac{(t+1)\ldots(t+n)}{t^{n}}} - \frac{1}{t} &amp;= {} \frac{1}{t} \sqrt[n]{(t+1)\ldots(t+n)} - \frac{1}{t} \\[2mm]
 &amp;= \frac{1}{t} \sqrt[n]{ n! + u(t) } - \frac{1}{t} \\[2mm]
\end{align*}
$$

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 !
There is a very simple trick that involves bounding such products by matching terms of opposite sizes. This idea motivates considering the following inequalities.

$(x+1)(x+n) \le (x+k)(x+n+1-k) \le (x+\frac{n+1}{2})(x+\frac{n+1}{2})$ for any $x \ge 0$ and $k \in [1..n]$.

The above can be proven by many different methods. After this, it is trivial to bound the expression under the limit by $\sqrt{(x+1)(x+n)} - x$ and $(x+\frac{n+1}{2}) - x$, and then taking limits for both yield $\frac{n+1}{2}$ as $x \to \infty$, and hence by squeeze theorem we are done.
$\bf{My\; Solution::}$ 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 $$(x+1),(x+2),(x+3),............(x+n)&gt;0\;,$$ 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 $$\displaystyle \lim_{n\rightarrow \infty}\sqrt[n]{(x+1)(x+2)...........(x+n)}-x = \frac{n+1}{2}$$