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 !
 A: 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.
A: $$ \lim \limits_{x \to +\infty} \sqrt[n]{(x+1)\ldots(x+n)} - x$$
$$=\lim_{t\to0}\frac{\sqrt[n]{(1+t)(1+2t)\cdots(1+nt)}-1}t$$
$$=\lim_{t\to0}\frac{(1+t)(1+2t)\cdots(1+nt)-1}t\frac1{\sum_{r=0}^{n-1}\lim_{t\to0}(\sqrt[n]{(1+t)(1+2t)\cdots(1+nt)})^r}$$
$$=\lim_{t\to0}\frac{(1+2+\cdots+n)t+O(t^2)}t\cdot\frac1{\sum_{r=0}^{n-1}1}$$
A: Asymptotic results
I proved
a number of years ago
(when I write
$a(n) \approx b(n)-c(n)$
I mean
$\lim_{n \to \infty} \dfrac{b(n)-a(n)}{c(n)}
=1
$):
$$
x(x+1)...(x+n-1)
\approx(x+(n-1)/2)^n
-(x+(n-1)/2)^{n-2}\dfrac{n^3-n}{24}
$$
and
$$
(x(x+1)...(x+n-1))^{1/n}
\approx(x+(n-1)/2)
-\dfrac{n^2-1}{24(x+(n-1)/2)}
$$
The second result above
improves the approximation
requested by the OP.
Here is a transcription 
from part of my PDF:
Let
$P_n(x)
=\prod_{i=0}^{n-1} (x+i)
$,
$r
=\frac{n-1}{2}
$
and
$u
=x+r
$.
Note that
$r$ is a function of $n$,
and
$u$ and $v$
(defined below)
are functions of
$n$ and $x$.
If
$S_i(n)
=\frac1{2i}\sum_{j=0}^{n-1} (r-j)^{2i}
$
then,
for any $k \ge 1$,
$$\sum_{i=1}^k \dfrac{S_i(n)}{u^{2i}}
\le \ln\left(\dfrac{u^n}{P_n(x)}\right)
\le \sum_{i=1}^{k-1} \dfrac{S_i(n)}{u^{2i}}
+\dfrac{S_k(n)}{u^{2k}(1-r^2/u^2}
$$
This follows from
$P_n^2(x)
=\prod_{j=0}^{n-1} (u^2-(r-j)^2)
$
and the result
valid for $k \ge 1$
and
$0 \le z < 1$
that
$\sum_{i=1}^k \dfrac{z^i}{i}
\le -\ln(1-z)
\le \sum_{i=1}^{k-1} \dfrac{z^i}{i}
+\dfrac{z^k}{k(1-z)}
$.
Explicit expressions are
$S_1(n)
=\dfrac{n^3-n}{24}
$
and
$S_2(n)
=\dfrac{(n^3-n)(3n^2-7)}{960}
$.
Letting $k=1$ in this result
and letting
$v
=(P_n(x))^{1/n}
$,
this becomes
$\dfrac{n^3-n}{24u^2}
\le \ln\left(\dfrac{u^n}{P_n(x)}\right)
\le \dfrac{n^3-n}{24(u^2-r^2)}
$
and
$\dfrac{n^2-1}{24u^2}
\le \ln\left(\dfrac{u}{v}\right)
\le \dfrac{n^2-1}{24(u^2-r^2)}
$.
To convert these inequalities involving logs 
to inequalities
using differences, 
we will use the following result.
If $c/u^2 \le  \ln(a/b) \le c/(u^2 - r^2)$
 where $a, b,$ and $c$
are all positive
and c are all positive, then
A. $a - b \ge bc/u^2$;
B: $a - b \le ac/(u^2 - r^2)$;
for any $d > 0$,
C:  if 
$u
\ge  r\sqrt{1 + 1/d}
$
 then 
$a - b 
\le ac(1 + d)/u^2
$;
D if 
$u
\ge  r\sqrt{1 + c/(r^2d)}
$ 
then 
$a - b
\ge  ac(1 - d)/u^2
$.
Proof. Follows from the inequality 
true for all $z > 0$ that
$1 - 1/z
\le  \ln z 
\le z - 1
$.
From this,
after a moderate amount of algebra,
we get these:
$$
\lim_{n \to \infty}\dfrac{u^n}{P_n(x)}
= \lim_{n \to \infty}\dfrac{u}{v}
=1
,
$$
$$
\lim_{n \to \infty}\dfrac{u^n-P_n(x)}{u^{n-2}}
= \dfrac{n^3-n}{24}
,
$$
and
$$
\lim_{n \to \infty} u(u-v)
=\dfrac{n^2-1}{24}
.
$$
These give the result
I quoted at the begining.
A: 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
$$\frac{n(n+1)/2 + O(1/x)}{n + O(1/x)} \to \frac{(n+1)}2 \text{ as }x \to \infty$$
A: $\bf{My\; Solution::}$ 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 $$\displaystyle \lim_{n\rightarrow \infty}\sqrt[n]{(x+1)(x+2)...........(x+n)}-x = \frac{n+1}{2}$$
